Topics in the Foundations of General Relativity and Newtonian Gravitation Theory

Topics in the Foundations of General Relativity and Newtonian Gravitation Theory (version 4, August 17, 2011) David B. ...

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Topics in the Foundations of General Relativity and Newtonian Gravitation Theory (version 4, August 17, 2011)

David B. Malament Department of Logic and Philosophy of Science University of California, Irvine [email protected] http://www.lps.uci.edu/lps bios/dmalamen

i

To Pen

Contents Preface

iv

1 Differential Geometry

1

1.1

Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1.2

Tangent Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8

1.3

Vector Fields, Integral Curves, and Flows . . . . . . . . . . . . . . . . . . . . . . . . . . .

16

1.4

Tensors and Tensor Fields on Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . .

22

1.5

The Action of Smooth Maps on Tensor Fields . . . . . . . . . . . . . . . . . . . . . . . . .

32

1.6

Lie Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

38

1.7

Derivative Operators and Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

44

1.8

Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

60

1.9

Metrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

65

1.10 Hypersurfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

82

1.11 Volume Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

97

2 Classical Relativity Theory

103

2.1

Relativistic Spacetimes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

2.2

Temporal Orientation and “Causal Connectibility” . . . . . . . . . . . . . . . . . . . . . . 111

2.3

Proper Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

2.4

Space/Time Decomposition at a Point and Particle Dynamics . . . . . . . . . . . . . . . . 122

2.5

The Energy-Momentum Field Tab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

2.6

Electromagnetic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

2.7

Einstein’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

2.8

Fluid Flow

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 ii

CONTENTS

2.9

iii

Killing Fields and Conserved Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

2.10 The Initial Value Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 2.11 Friedmann Spacetimes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 3 Special Topics

170

3.1

G¨ odel Spacetime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

3.2

Two Criteria of Orbital (Non-)Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

3.3

A No-Go Theorem about Orbital (Non-)Rotation . . . . . . . . . . . . . . . . . . . . . . . 206

4 Newtonian Gravitation Theory

215

4.1

Classical Spacetimes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216

4.2

Geometrized Newtonian Theory — First Version . . . . . . . . . . . . . . . . . . . . . . . 231

4.3

Interpreting the Curvature Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

4.4

A Solution to an Old Problem about Newtonian Cosmology . . . . . . . . . . . . . . . . . 249

4.5

Geometrized Newtonian Theory — Second Version . . . . . . . . . . . . . . . . . . . . . . 256

Solutions to Problems

265

Bibliography

292

Preface This manuscript began life as a set of lecture notes for a two-quarter (20 week) course on the foundations of general relativity that I taught at the University of Chicago many years ago. I have repeated the course quite a few times since then, both there and at the University of California, Irvine, and have over the years steadily revised the notes and added new material. Maybe now the notes can stand on their own. The course was never intended to be a systematic survey of general relativity. There are many standard topics that I do not discuss, e.g., the Schwarzschild solution and the “classic tests” of general relativity. (And I have always recommended that students who have not already taken a more standard course in the subject do some additional reading on their own.) My goals instead have been to (i) present the basic logical-mathematical structure of the theory with some care, and (ii) consider additional special topics that seem to me, at least, of particular interest. The topics have varied from year to year, and not all have found their way into these notes. I will mention in advance three that did. The first is “geometrized Newtonian gravitation theory”, also known as “Newton-Cartan theory”. It is now well known that one can, after the fact, reformulate Newtonian gravitation theory so that it exhibits many of the qualitative features that were once thought to be uniquely characteristic of general relativity. On reformulation, Newtonian theory too provides an account of four-dimensional spacetime structure in which (i) gravity emerges as a manifestation of spacetime curvature, and (ii) spacetime structure itself is “dynamical” in the sense that it participates in the unfolding of physics rather than being a fixed backdrop against which it unfolds. It has always seemed to me helpful to consider general relativity and this geometrized reformulation of Newtonian theory side by side. For one thing, one derives a sense of where Einstein’s equation “comes from”. When one reformulates the empty-space field equation of Newtonian gravitation theory (i.e., Laplace’s equation ∇2 φ = 0, where φ is the gravitational potential),

one arrives at a constraint on the curvature of spacetime, namely Rab = 0. The latter is, of course,

just what we otherwise know as (the empty-space version of) Einstein’s equation. And, reciprocally, this comparison of the two theories side by side provides a certain insight into Newtonian physics. For example, it yields a satisfying solution (or dissolution) to an old problem about Newtonian cosmology. Newtonian theory in a standard textbook formulation seems to provide no sensible prescription for what the gravitational field should be like in the presence of a uniform mass-distribution filling all of space. iv

PREFACE

v

(See section 4.4.) But the problem is really just an artifact of the formulation, and it disappears when one passes to the geometrized version of the theory. The basic idea of geometrized Newtonian gravitation theory is simple enough. But there are complications, and I deal with some of them in the present expanded form of the lecture notes. In particular, I present two different versions of the theory – what I call the “Trautman version” and the “K¨ unzle-Ehlers version” – and consider their relation to one another. I also discuss in some detail the geometric significance of various conditions on the Riemann curvature field Ra bcd that enter into the formulation of these versions. A second special topic that I consider is the concept of “rotation”. It turns out to be a rather delicate and interesting question, at least in some cases, just what it means to say that a body is or is not rotating within the framework of general relativity. Moreover, the reasons for this — at least the ones I have in mind — do not have much to do with traditional controversy over “absolute vs. relative (or Machian)” conceptions of motion. Rather they concern particular geometric complexities that arise when one allows for the possibility of spacetime curvature. The relevant distinction for my purposes is not that between attributions of “relative” and “absolute” rotation, but rather that between attributions of rotation that can and cannot be analyzed in terms of motion (in the limit) at a point. It is the latter — ones that make essential reference to extended regions of spacetime — that can be problematic. The problem has two parts. First, one can easily think of different criteria for when an extended body is rotating. (I discuss two examples in section 3.2.) These criteria agree if the background spacetime structure is sufficiently simple, e.g., if one is working in Minkowski spacetime. But they do not agree in general. So, at the very least, attributions of rotation in general relativity can be ambiguous. A body can be rotating in one perfectly natural sense but not rotating in another, equally natural, sense. Second, circumstances can arise in which the different criteria – all of them – lead to determinations of rotation and non-rotation that seem wildly counterintuitive. (See section 3.3.) The upshot of this discussion is not that we cannot continue to talk about rotation in the context of general relativity. Not at all. Rather, we simply have to appreciate that it is a subtle and ambiguous notion that does not, in all cases, fully answer to our classical intuitions. A third special topic that I consider is Gödel spacetime. It is not a live candidate for describing our universe, but it is of interest because of what it tells us about the possibilities allowed by general relativity. It represents a possible universe with remarkable properties. For one thing, the entire material content of the G¨ odel universe is in a state of uniform, rigid rotation (according to any reasonable criterion of rotation). For another, light rays and free test particles in it exhibit a kind of boomerang effect. Most striking of all, it admits closed timelike curves that cannot be “unrolled” by passing to a covering space (because the underlying manifold is simply connected). In section 3.1, I review these basic features of Gödel spacetime and, in an appendix to that section, discuss how one can go back and forth between an intrinsic characterization of the G¨ odel metric and two different coordinate expressions for it.

PREFACE

vi

These three special topics are treated in chapters 3 and 4. Much of this material has been added over the years. The original core of the lecture notes — the review of the basic structure of general relativity — is to be found in chapter 2. Chapter 1 offers a preparatory review of basic differential geometry. It has never been my practice to work through all this material in class. I have limited myself there to “highlights” and general remarks. But I have always distributed the notes so that students with sufficient interest can do further reading on their own. On occasion, I have also run a separate “problem session” and used it for additional coaching on differential geometry. (A number of problems, with solutions, are included in the present version of the notes.) I suggest that readers make use of chapter 1 as seems best to them — as a text to be read from the beginning, as a reference work to be consulted when particular topics arise in later chapters, as something in between, or not at all. I would like to use this occasion to thank a number of people who have helped me over the years to learn and better understand general relativity. I could produce a long list, but the ones who come first, at least, are John Earman, David Garfinkle, Robert Geroch, Clark Glymour, Howard Stein, and Robert Wald. I am particularly grateful to Bob1 and Bob2 for allowing this interloper from the Philosophy Department to find a second home in the Chicago Relativity Group. Anyone familiar with their work, both research and expository writings, will recognize their influence on this set of lecture notes. Erik Curiel, Sam Fletcher, David Garfinkle, John Manchak, and Jim Weatherall have my thanks, as well, for the comments and corrections they have given me on earlier drafts of the manuscript. Matthias Kretschmann was good enough some years ago to take my handwritten notes on differential geometry and set them in TEX. I took over after that, but I might not have started without his push. Finally, Pen Maddy has helped me to believe that this project was worth completing. I shall always be grateful to her for her support and encouragement.

Chapter 1

Differential Geometry 1.1

Manifolds

We assume familiarity with the basic elements of multivariable calculus and point set topology. The following notions, in particular, should be familiar. Rn (for n ≥ 1) is the set of all n-tuples of real numbers x = (x1 , ..., xn ). The Euclidean inner product √ (or “dot product”) on Rn is given by x · y = x1 y 1 + ... + xn y n . It determines a norm kxk = x · x.

Given a point x ∈ Rn and a real number ǫ > 0, Bǫ (x) is the open ball in Rn centered at x with radius

ǫ, i.e., Bǫ (x) = {y : ky − xk < ǫ}. Clearly, x belongs to Bǫ (x) for every ǫ > 0. A subset S of Rn is open

if, for all points x in S, there is an ǫ > 0 such that Bǫ (x) ⊆ S. This determines a topology on Rn . Given

m, n ≥ 1, and a map f : O → Rm from an open set O in Rn to Rm , f is smooth (or C ∞ ) if all its mixed

partial derivatives (to all orders) exist and are continuous at every point in O.

A smooth n-dimensional manifold (n ≥ 1) can be thought of as a point set to which has been added the

“local smoothness structure” of Rn . Our discussion of differential geometry begins with a more precise characterization.1 Let M be a non-empty set. An n-chart on M is a pair (U, ϕ) where U is a subset of M and ϕ : U → Rn

is an injective (i.e., one-to-one) map from U into Rn with the property that ϕ[U ] is an open subset of Rn . (Here ϕ[U ] is the image set {ϕ(p) : p ∈ U }.) Charts, also called “coordinate patches”, are the mechanism

with which one induces local smoothness structure on the set M . To obtain a smooth n-dimensional manifold, we must lay down sufficiently many n-charts on M to cover the set, and require that they are, in an appropriate sense, compatible with one another. Let (U1 , ϕ1 ) and (U2 , ϕ2 ) be n-charts on M . We say the two are compatible if either the intersection set U = U1 ∩ U2 is empty, or the following conditions hold: 1 In

this section and several others in chapter 1, we follow the basic lines of the presentation in Geroch [22].

1

2

CHAPTER 1. DIFFERENTIAL GEOMETRY

(1) ϕ1 [U ] and ϕ2 [U ] are both open subsets of Rn ; n n (2) ϕ1 ◦ ϕ−1 and ϕ2 ◦ ϕ−1 are both smooth. 2 : ϕ2 [U ] → R 1 : ϕ1 [U ] → R

(Notice that the second makes sense since ϕ1 [U ] and ϕ2 [U ] are open subsets of Rn and we know what it means to say that a map from an open subset of Rn to Rn is smooth. See figure 1.1.1.) The relation of compatibility between n-charts on a given set is reflexive and symmetric. But it need not be transitive and, hence, not an equivalence relation. For example, consider the following three 1−charts on R: C1 = (U1 , ϕ1 )

with U1 = (−1, 1)

and ϕ1 (x) = x

C2 = (U2 , ϕ2 )

with U2 = (0, 1)

and ϕ2 (x) = x

C3 = (U3 , ϕ3 )

with U3 = (−1, 1)

and ϕ3 (x) = x3 .

Pairs C1 and C2 are compatible, and so are pairs C2 and C3 . But C1 and C3 are not compatible, because the map ϕ1 ◦ ϕ−1 3 : (−1, +1) → R is not smooth (or even just differentiable) at x = 0.

M

ϕ1

Rn ϕ1 [U1 ]

U1 ϕ2 U2

ϕ2 [U2 ]

Figure 1.1.1: Two n-charts (U1 , ϕ1 ) and (U2 , ϕ2 ) on M with overlapping domains.

We now define a smooth n-dimensional manifold (or, in brief, an n-manifold ) (n ≥ 1) to be a pair

(M, C) where M is a non-empty set, and C is a set of n-charts on M satisfying the following four conditions.

(M1) Any two n-charts in C are compatible. (M2) The (domains of the) n-charts in C cover M , i.e., for every p ∈ M , there is an n-chart (U, ϕ) in C such that p ∈ U .

(M3) (Hausdorff condition) Given distinct points p1 and p2 in M , there exist n-charts (U1 , ϕ1 ) and (U2 , ϕ2 ) in C such that pi ∈ Ui for i = 1, 2, and U1 ∩ U2 is empty. (M4) C is maximal in the sense that any n-chart on M that is compatible with every n-chart in C belongs to C.

3


(M1) and (M2) are certainly conditions one would expect. (M3) is included, following standard practice, simply to rule out pathological examples (though one does, sometimes, encounter discussions of “nonHausdorff manifolds”). (M4) builds in the requirement that manifolds don’t have “extra structure” in the form of distinguished n-charts. (For example, we can think of the point set Rn as carrying a single (global) n-chart. In the transition from the point set Rn to the n-manifold Rn discussed below, this “extra structure” is washed out.) Because of (M4), it might seem a difficult task to specify an n-dimensional manifold. (How is one to get a grip on all the different n-charts that make up a maximal set of such?) But the following proposition shows that the specification need not be difficult. It suffices to come up with a set of n-charts on the underlying set satisfying (M1), (M2), (M3), and then simply throw in wholesale all other compatible n-charts. Proposition 1.1.1. Let M be a non-empty set, let C0 be a set of n-charts on M satisfying conditions (M 1), (M 2), (M 3), and let C be the set of all n-charts on M compatible with all the n-charts in C0 . Then (M, C) is an n-manifold, i.e., C satisfies all four conditions.

Proof. Since C0 satisfies (M1), C0 is a subset of C. It follows immediately that C satisfies (M2), (M3), and

(M4). Only (M1) requires some argument. Let C1 = (U1 , ϕ1 ) and C2 = (U2 , ϕ2 ) be any two n-charts

compatible with all n-charts in C0 . We show that they are compatible with one another. We may assume that the intersection U1 ∩ U2 is non-empty, since otherwise compatibility is automatic.

First we show that ϕ1 [U1 ∩U2 ] is open. (A parallel argument establishes that ϕ2 [U1 ∩U2 ] is open.)

Consider an arbitrary point of ϕ1 [U1 ∩ U2 ]. It is of the form ϕ1 (p) for some point p ∈ U1 ∩ U2 . Since C0

satisfies (M2), there exists an n-chart C = (U, ϕ) in C0 whose domain contains p. So p ∈ U ∩ U1 ∩ U2 .

Since C is compatible with both C1 and C2 , ϕ[U ∩ U1 ] and ϕ[U ∩ U2 ] are open sets in Rn , and the

maps

ϕ1 ◦ ϕ−1 : ϕ[U ∩ U1 ] → Rn ,

ϕ2 ◦ ϕ−1 : ϕ[U ∩ U2 ] → Rn ,

n ϕ ◦ ϕ−1 1 : ϕ1 [U ∩ U1 ] → R ,

n ϕ ◦ ϕ−1 2 : ϕ2 [U ∩ U2 ] → R ,

are all smooth (and so continuous). Now ϕ[U ∩ U1 ∩ U2 ] is open, since it is the intersection of open sets ϕ[U ∩ U1 ] and ϕ[U ∩ U2 ].

(Here we use the fact that ϕ is injective.) So ϕ1 [U ∩ U1 ∩ U2 ] is

open, since it is the pre-image of ϕ[U ∩ U1 ∩ U2 ] under the continuous map ϕ ◦ ϕ−1 1 .

But, clearly,

ϕ1 (p) ∈ ϕ1 [U ∩ U1 ∩ U2 ], and ϕ1 [U ∩ U1 ∩ U2 ] is a subset of ϕ1 [U1 ∩ U2 ]. So we see that our arbitrary

point ϕ1 (p) in ϕ1 [U1 ∩ U2 ] is contained in an open subset of ϕ1 [U1 ∩ U2 ]. Thus ϕ1 [U1 ∩ U2 ] is open. n Next we show that the map ϕ2 ◦ ϕ−1 is smooth. (A parallel argument establishes 1 : ϕ1 [U1 ∩ U2 ] → R

n that ϕ1 ◦ ϕ−1 is smooth.) For this it suffices to show that, given our arbitrary point 2 : ϕ2 [U1 ∩ U2 ] → R

ϕ1 (p) in ϕ1 [U1 ∩ U2 ], the restriction of ϕ2 ◦ ϕ1−1 to some open subset of ϕ1 [U1 ∩ U2 ] containing ϕ1 (p)

is smooth. But this follows easily. We know that ϕ1 [U ∩ U1 ∩ U2 ] is an open subset of ϕ1 [U1 ∩ U2 ]

to ϕ1 [U ∩ U1 ∩ U2 ] is smooth, since it can be containing ϕ1 (p). And the restriction of ϕ2 ◦ ϕ−1 1


4

realized as the composition of ϕ ◦ ϕ−1 (restricted to ϕ1 [U ∩ U1 ∩ U2 ]) with ϕ2 ◦ ϕ−1 (restricted to 1 ϕ[U ∩ U1 ∩ U2 ]), and both these maps are smooth.

Our definition of manifolds is less restrictive than some in that we do not include the following condition. (M5) (Countable cover condition) There is a countable subset {(Un , ϕn ) : n ∈ N} of C whose domains cover M , i.e., for all p in M , there is an n such that p ∈ Un .

In fact, all the manifolds that one encounters in relativity theory satisfy (M5). But there is some advantage in not taking the condition for granted from the start. It is simply not needed for our work until we discuss derivative operators, i.e., affine connections, on manifolds in section 1.7. It turns out that (M5) is actually a necessary and sufficient condition for there to exist a derivative operator on a manifold (given our characterization). It is also a necessary and sufficient condition for there to exist a (positive definite) Riemannian metric on a manifold. (See Geroch [23]. The paper gives a nice example of a 2-manifold that violates (M5).) Our way of defining n-manifolds is also slightly non-standard because we jump directly from the point set M to the manifold (M, C). In contrast, one often proceeds in two stages. One first puts a topology T

on M forming a topological space (M, T ). Then one adds the set of n-charts C to form the “manifold” (M, T ), C . If one proceeds this way, one must require of every n-chart (U, ϕ) in C that U be open, i.e., that U belong to T , so that ϕ : U → Rn qualifies as continuous.

Given our characterization of an n-manifold (M, C), we do not (yet) know what it means for a subset

of M to be “open”. But there is a natural way to use the n-charts in C to define a topology on M . We

say that a subset S of M is open if, for all p in S, there is an n-chart (U, ϕ) in C such that p ∈ U and

U ⊆ S. (This topology can also be characterized as the coarsest topology on M with respect to which, for all n-charts (U, ϕ) in C, ϕ : U → Rn is continuous. See problem 1.1.3). It follows immediately that

the domain of every n-chart is open.

b be an open subset Problem 1.1.1. Let (M, C) be an n-manifold, let (U, ϕ) be an n-chart in C, let O −1 b of ϕ[U ], and let O be its preimage ϕ [O]. (So, O ⊆ U .) Show that (O, ϕ|O ), the restriction of (U, ϕ) to O, is also an n-chart in C.

Problem 1.1.2. Let (M, C) be an n-manifold, let (U, ϕ) be an n-chart in C, and let O be an open set in M such that U ∩ O 6= ∅. Show that U ∩ O, ϕ|U∩O , the restriction of (U, ϕ) to U ∩ O, is also an n-chart in C. (Hint: Make use of the result in problem 1.1.1. Strictly speaking, by the way, we do not need to assume that U ∩ O is non-empty. But that is the only case of interest.)

Problem 1.1.3. Let (M, C) be an n-manifold and let T be the set of open subsets of M . (i) Show that

T is, in fact, a topology on M , i.e., it contains the empty set and the set M , and is closed under finite


5

intersections and arbitrary unions. (ii) Show that T is the coarsest topology on M with respect to which ϕ : U → Rn is continuous for all n-charts (U, ϕ) in C.

Now we consider a few examples of manifolds. Let M be Rn , the set of all ordered n-tuples of real numbers. Let U be any subset of M that is open (in the standard topology on Rn ), and let ϕ : U → Rn

be the identity map. Then (U, ϕ) qualifies as an n-chart on M . Let C0 be the set of all n-charts on M of

this very special form. It is easy to check that C0 satisfies conditions (M1), (M2), (M3). If we take C to

be the set of all n-charts on M compatible with all n-charts in C0 , then it follows (by proposition 1.1.1)

that (M, C) is an n-manifold. We refer to it as “the manifold Rn ”. (Thus, one must distinguish among the point set Rn , the vector space Rn , the manifold Rn , and so forth.)

Next we introduce the manifold S n . The underlying set M is the set of points x = (x1 , ..., xn+1 ) ∈

Rn+1 such that kxk = 1. For each i = 1, ..., n + 1, we set Ui+

=

{(x1 , ..., xi , ..., xn+1 ) ∈ M : xi > 0},

Ui−

=

{(x1 , ..., xi , ..., xn+1 ) ∈ M : xi < 0},

+ − n n and define maps ϕ+ and ϕ− by setting i : Ui → R i : Ui → R 1 n+1 1 n+1 ϕ+ ) = (x1 , ..., xi−1 , xi+1 , ..., xn+1 ) = ϕ− ). i (x , ..., x i (x , ..., x − Ui+ and Ui− are upper and lower hemispheres with respect to the xi coordinate axis; ϕ+ i and ϕi are projections that erase the ith coordinate of (x1 , ..., xn+1 ). The (n + 1) pairs of the form (Ui+ , ϕ+ i )

and (Ui− , ϕ− i ) are n-charts on M . The set C1 of all such pairs satisfies conditions (M1) and (M2). For

all p ∈ M and all ǫ > 0, if Bǫ (p) ∩ M is a subset of Ui+ (respectively Ui− ), we now add to C1 the

− − n-chart that results from restricting (Ui+ , ϕ+ i ) (respectively (Ui , ϕi )) to Bǫ (p) ∩ M . The expanded

set of n-charts C2 satisfies (M1), (M2), (M3). If, finally, we add to C2 all n-charts on M compatible with all n-charts in C2 , we obtain the n-manifold S n .

We thus have the manifolds Rn and S n for every n ≥ 1. From these we can generate many more

manifolds by taking products and cutting holes.

Let M1 = (M1 , C1 ) be an n1 -manifold and let M2 = (M2 , C2 ) be an n2 -manifold. The product

manifold M1 × M2 is an (n1 + n2 )−manifold defined as follows. The underlying point set is just the

Cartesian product M1 × M2 , i.e., the set of all pairs (p1 , p2 ) where pi ∈ Mi for i = 1, 2. Let (U1 , ϕ1 )

be an n1 -chart in C1 and let (U2 , ϕ2 ) be an n2 -chart in C2 . We associate with them a set U and a map ϕ : U → R(n1 +n2 ) . We take U to be the product U1 × U2 ; and given (p1 , p2 ) ∈ U , we take ϕ (p1 , p2 ) to be (y 1 , ..., y n1 , z 1 , ..., z n2 ), where ϕ1 (p1 ) = (y 1 , ..., y n1 ), and ϕ2 (p2 ) = (z 1 , ..., z n2 ).

So defined,

(U, ϕ) qualifies as an (n1 + n2 )-chart on M1 × M2 . The set of all (n1 + n2 )-charts on M1 × M2 obtained in this manner satisfies conditions (M1), (M2), (M3). If we now throw in all n-charts on M1 × M2

that are compatible with all members of this set, we obtain the manifold M1 × M2 . Using this product


6

construction, we generate the 2−manifold R1 ×S 1 (the “cylinder”), the 2−manifold S 1 ×S 1 (the “torus”),

and so forth.

Next, let (M, C) be an n-manifold, and let S be a closed proper subset of M . (So M − S is a

non-empty open subset of M .) Further, let C ′ be the set of all n-charts (U, ϕ) in C where U ⊆ (M − S).

Then the pair (M − S, C ′ ) is an n-manifold in its own right. (This follows as a corollary to the assertion

in problem 1.1.2.)

A large fraction of the manifolds one encounters in relativity theory can be obtained from the manifolds n

R and S n by taking products and excising closed sets. We now define “smooth maps” between manifolds. We do so in two stages. First, we consider the special case in which the second manifold (i.e., the one into which the first is mapped) is R. Then we consider the general case. Let (M, C) be an n-manifold. We say that a map α : M → R is smooth (or

C ∞ ) if, for all n-charts (U, ϕ) in C, α ◦ ϕ−1 : ϕ[U ] → R is smooth. (Here we use a standard technique.

To define something on an n-manifold we use the charts to pull things back to the context of Rn where the notion already makes sense.) Next let (M ′ , C ′ ) be an m-manifold (with no requirement that m = n).

We say that a map ψ : M → M ′ is smooth (or C ∞ ) if, for all smooth maps α : M ′ → R on the second manifold, the composed map α ◦ ψ : M → R is smooth. One can check that the second definition is

compatible with the first (see problem 1.1.4), and with the standard definition of smoothness that applies

specifically to maps of the form ψ : Rn → Rm . Problem 1.1.4. Let (M, C) be an n-manifold. Show that a map α : M → R is smooth according to our

first definition (which applies only to real-valued maps on manifolds) iff it is smooth according to our second definition (which applies to maps between arbitrary manifolds). Let (M, C) and (M ′ , C ′ ) be manifolds. The definition of smoothness just given naturally extends to

maps of the form ψ : O → M ′ where O is an open subset of M (that need not be all of M ). It does so

because we can always think of O as a manifold in its own right when paired with the charts it inherits from C, i.e., the charts in C whose domains are subsets of O. On this understanding it follows, for example, that if a map ψ : M → M ′ is smooth, then its restriction to O is smooth. It also follows that given any

chart (U, ϕ) in C, the maps ϕ : U → Rn and ϕ−1 : ϕ[U ] → M are both smooth.

The point mentioned in the preceding paragraph will come up repeatedly. We shall often formulate definitions in terms of structures defined on manifolds and then transfer them without comment to open subsets of manifolds. It should be understood in each case that we have in mind the manifold structure induced on those open sets. Given manifolds (M, C) and (M ′ , C ′ ), a bijection ψ : M → M ′ is said to be a diffeomorphism if

both ψ and ψ −1 are smooth. Two manifolds are said to be diffeomorphic, of course, if there exists a diffeomorphism between them, i.e., between their underlying point sets. Diffeomorphic manifolds are


7

as alike as they can be with respect to their “structure”. They can differ only in the identity of their underlying elements.

8


1.2

Tangent Vectors

Let (M, C) be an n-manifold and let p be a point in M . In this section, we introduce the notion of a

“vector” (or “tangent vector” or “contravariant vector”) at p. We also show that the set of all vectors at p naturally forms an n-dimensional vector space. Consider first the familiar case of Rn . A vector ξ at a point in Rn can be characterized by its components (ξ 1 , ..., ξ n ) with respect to the n coordinate axes. This characterization is not available for arbitrary n-manifolds where no coordinate curves are distinguished. But an alternate, equivalent characterization does lend itself to generalization. Let p be a point in Rn . We take S(p) to be the set of all smooth maps f : O → R, where O is some

open subset (or other) of Rn that contains p. If f1 : O1 → R and f2 : O2 → R are both in S(p),

then we can define new maps (f1 + f2 ) : O1 ∩ O2 → R and (f1 f2 ) : O1 ∩ O2 → R in S(p) by setting (f1 + f2 )(q) = f1 (q) + f2 (q) and (f1 f2 )(q) = f1 (q) f2 (q) for all points q in O1 ∩ O2 .

Now suppose that ξ is a vector at p in Rn with components (ξ 1 , ..., ξ n ) and that f is in S(p). The

directional derivative of f at p in the direction ξ is defined by: ξ(f ) = ξ · (∇f )|p =

n X i=1

ξi

∂f (p). ∂xi

(1.2.1)

It follows immediately from the elementary properties of partial derivatives that, for all f1 and f2 in S(p), (DD1) ξ(f1 + f2 ) = ξ(f1 ) + ξ(f2 ) (DD2) ξ(f1 f2 ) = f1 (p) ξ(f2 ) + f2 (p) ξ(f1 ) (DD3) If f1 is constant, ξ(f1 ) = 0. Any map from S(p) to R satisfying these three conditions will be called a derivation (or directional

derivative operator ) at p. Thus, every vector at p defines, via (1.2.1), a derivation at p. Indeed, we shall see in a moment that (1.2.1) defines a bijection between vectors at p (understood as ordered n-tuples of reals) and derivations at p. This will give us our desired alternate characterization of vectors in Rn . But first we need a lemma. Lemma 1.2.1. Let f1 : O1 → R and f2 : O2 → R be elements of S(p) that agree on some open set O ⊆ O1 ∩ O2 containing p. Then, for all derivations ξ at p, ξ(f1 ) = ξ(f2 ).

Proof. Let h : O → R be the constant map on O that assigns 1 to all points. Certainly h is in S(p). The

maps h f1 and h f2 have domain O and agree throughout O, i.e., h f1 = h f2 . So ξ(h f1 ) = ξ(h f2 ). But by (DD2) and (DD3), ξ(h f1 ) = h(p) ξ(f1 ) + f1 (p) ξ(h) = 1 ξ(f1 ) + f1 (p) 0 = ξ(f1 ). Similarly, ξ(h f2 ) = ξ(f2 ). So ξ(f1 ) = ξ(f2 ), as claimed.

9


Proposition 1.2.2. Equation (1.2.1) defines a bijection between vectors at p and derivations at p. Proof. Suppose first that ξ = (ξ 1 , ..., ξ n ) and η = (η 1 , ..., η n ) are vectors at p that, via (1.2.1), determine the same derivation at p. Then ξ · (∇f )|p = η · (∇f )|p , for all f in S(p). Consider the special case where

f is the coordinate map xi : Rn → R that assigns to a point in Rn its ith coordinate. We have (∇xi )|p =

∂xi ∂xi ∂xi , ..., , ..., = (0, ..., 0, 1, 0, ..., 0), ∂x1 ∂xi ∂xn p

where the sole 1 in the far right n-tuple is in the ith position. So ξ i = ξ · (∇xi )|p = η · (∇xi )|p = η i . But this is true for all i = 1, ..., n. Hence ξ = η. Thus, the map from vectors at p to derivations at p determined by (1.2.1) is injective. Next, suppose that ξ is a derivation at p and that the numbers ξ 1 , ..., ξ n are defined by ξ i = ξ(xi ). n X ∂f ξ i i (p). That is, we show that ξ can be realized as We show that, for all f in S(p), ξ(f ) = ∂x i=1

the image of (ξ 1 , ..., ξ n ) under the map determined by (1.2.1). This will establish that the map is also

surjective. Let f : O → R be a map in S(p). By the preceding lemma, we may assume that O is an open ball

centered at p. (If f ′ is the restriction of f to an open ball centered at p, ξ(f ′ ) = ξ(f ).

So we lose

′

nothing by working with f rather than f .) If x is a point in O, it follows by the “fundamental theorem of calculus” that f (x) = f (p) +

Z

1

0

d f p + t(x − p) dt. dt

(We want the domain of f to be an open ball centered at p to insure that f is defined at all points on the line segment connecting p and x.) By the “chain rule”, n X d ∂f (xi − pi ). f p + t(x − p) = p + t(x − p) i dt ∂x i=1 Inserting the right side of this equation into the integrand above, we arrive at f (x) = f (p) +

n X i=1

gi (x) (xi − pi ),

where, for all i, the map gi : O → R is given by gi (x) = S(p). It now follows from (DD1), (DD2) and (DD3) that ξ(f ) =

n h X i=1

Z

0

1

(1.2.2)

∂f (p + t(x − p)) dt. The gi belong to ∂xi

i gi (p) ξ(xi − pi ) + (xi − pi )(p) ξ(gi ) .

(Here we are construing the numbers f (p) and p1 , ..., pn as constant functions on O.) But (xi − pi )(p) =

pi − pi = 0, and ξ(xi − pi ) = ξ(xi ) − ξ(pi ) = ξ i − 0 = ξ i . So we have ξ(f ) =

n X i=1

ξ i gi (p).

10


n

X ∂f ∂f ξ i i (p), as claimed. But it follows from (1.2.2) that (p) = g (p). So ξ(f ) = i ∂xi ∂x i=1 With proposition 1.2.2 as motivation, we now give our definition of “vectors” at points of manifolds. Given a manifold (M, C) and a point p in M , let S(p) be the set of smooth maps f : O → R where O

is some open subset (or other) of M that contains p. (Our prior remark about adding and multiplying elements of S(p) carries over intact.) We take a vector (or tangent vector, or contravariant vector ) at p

to be a map from S(p) to R that satisfies (DD1), (DD2), and (DD3).

The set of all vectors at p has a natural vector space structure (over R). If ξ and η are vectors at p, and k is a real number, we can define new vectors ξ + η and k ξ by setting (ξ + η)(f )

= ξ(f ) + η(f ),

(k ξ)(f )

= k ξ(f ),

for all f in S(p). The vector space Mp so defined is call the tangent space to p. We shall soon show that

Mp has dimension n, i.e., has the same dimension as (M, C). To do so, we give a second characterization

of vectors on manifolds that is of independent interest.

Let γ : I → M be a smooth curve in M , i.e., a smooth map from an open interval I ⊆ R into M . (I

is of the form (a, b), (−∞, b), (a, +∞), or (−∞, +∞), where a and b are real numbers. We know what it means to say that γ : I → M is smooth since, as noted toward the end of section 1.1, we can think of

I as a manifold in its own right when paired with the charts it inherits from the manifold R.) Suppose → → d f ◦ γ (s0 ) for all s0 ∈ I and γ(s0 ) = p. We associate with γ a vector γ p at p by setting γ p (f ) = ds f in S(p). (This definition makes sense since (f ◦ γ) is a smooth map from I into R.) It is easy to check →

that γ p , so defined, satisfies (DD1) – (DD3). For example, (DD2) holds for all f1 and f2 in S since → d d γ p (f1 f2 ) = (f1 f2 ) ◦ γ (s0 ) = (f1 ◦ γ) (f2 ◦ γ) (s0 ) ds ds d d f2 ◦ γ (s0 ) + f2 ◦ γ (s0 ) f1 ◦ γ (s0 ) = f1 ◦ γ (s0 ) ds ds = → γp

→

→

f1 (p) γ p (f2 ) + f2 (p) γ p (f1 ).

is called the tangent vector to γ at p.

Suppose now that (U, ϕ) is an n-chart in our n-manifold (M, C). Associated with (U, ϕ) are coordinate maps ui : U → R for i = 1, ..., n defined by ui (q) = xi ◦ ϕ (q). Thus, the number that ui assigns to a point q in M is the one that xi assigns to the image point ϕ(q) in Rn . Equivalently, ui (q) is the ith coordinate of ϕ(q). So ϕ(q) = u1 (q), ..., un (q) . Now let p be a point in U . We understand the ith coordinate curve through ϕ(p) = u1 (p), ..., un (p)

in Rn to be the map from R to Rn given by

s 7→ u1 (p), ..., ui−1 (p), ui (p) + s, ui+1 (p), ..., un (p) .

(1.2.3)

11


M

Rn

ϕ

γj

p

ϕ(p)

γi U

ϕ[U ]

Figure 1.2.1: Coordinate curves on M with respect to (U, ϕ).

The image of the curve is a line through ϕ(p), parallel to the ith coordinate axis through the origin (see figure 1.2.1). We can pull this curve back to U via ϕ−1 to obtain a smooth curve γi : I → U through p: γi (s) = ϕ−1 u1 (p), ..., ui−1 (p), ui (p) + s, ui+1 (p), ..., un (p) .

(1.2.4)

Note that γi (0) = p. (We can afford to be vague about the domain I of γi since we are interested only in the tangent to the curve at p. All that matters is that 0 ∈ I. How do we know that γi is smooth?

This follows because ϕ−1 is smooth, and so γi is the composition of two smooth maps.) Extending our previous usage, we now refer to γi as the ith coordinate curve through p with respect to (U, ϕ). (Note

that coordinate curves through points in Rn are defined outright, but coordinate curves through points →

in M are necessarily relativized to n-charts.) This curve has a tangent γ i |p at p. By the chain rule, ! → ∂ f ◦ ϕ−1 d γ i |p (f ) = f ◦ γi (0) = ϕ(p) (1.2.5) i ds ∂x

for all f in S(p). We note for future reference, in particular, that since uj = xj ◦ ϕ, j → ∂x γ i |p (uj ) = ϕ(p) = δij . i ∂x

(1.2.6)

(Here δij is the Kronecker delta function that is 1 if i = j, and 0 otherwise.) Sometimes the tangent → → ∂f ∂ , and γ i (f ) is written as . Using this notation, and suppressing the point vector γ i is written as ∂ui ∂ui of evaluation p, (1.2.5) and (1.2.6) come out as

and

∂f ∂(f ◦ ϕ−1 ) = ∂ui ∂xi

(1.2.7)

∂uj = δij . ∂ui

(1.2.8)

→

Using the tangent vectors γ i |p , i = 1, ..., n, we can show that Mp is n-dimensional. Proposition 1.2.3. Let (M, C) be an n-manifold, let (U, ϕ) be an n-chart in C, let p be a point in U , and let γ1 , ..., γn be the n coordinate curves through p with respect to (U, ϕ). Then their tangent vectors → → γ 1 |p , ..., γ n |p

at p form a basis for Mp .

12


Proof. First we show that the vectors are linearly independent. Let a1 , ..., an be real numbers such that n → P ai γ i |p = 0. We must show that the ai are all 0. Now for all f in S(p), we have i=1

0=

n X

→ ai γ i |p

i=1

!

(f ) =

n X

→

ai γ i |p (f ).

i=1

→

Consider the special case where f is the coordinate map uj = xj ◦ ϕ on U . Then, by (1.2.6), γ i |p (f ) =

δij . So the equation reduces to 0 = aj . And this is true for all j = 1, ..., n.

Next, suppose that ξ is a vector at p. We show that it can be expressed as a linear combination of the First we associate with ξ a vector ξˆ at ϕ(p). (In what follows, we shall be going back and forth

→ γ i |p .

between the context of M and Rn . To reduce possible confusion, we shall systematically use carets for ˆ→R denoting objects associated with Rn ). We take ξˆ to be the vector whose action on elements fˆ: O ˆ fˆ) = ξ fˆ ◦ ϕ . This makes sense since fˆ ◦ ϕ is an element of S(p) with in S ϕ(p) is given by ξ( ˆ . By proposition 1.2.2 (applied to ξˆ at ϕ(p)), we know that there are real numbers domain ϕ−1 ϕ[U ]∩ O ξ 1 , ..., ξ n such that

ˆ fˆ) = ξ(

n X i=1

ξi

∂ fˆ ϕ(p) ∂xi

for all fˆ in S ϕ(p) . Now let f : O → R be an arbitrary element of S(p). Then f ◦ ϕ−1 : ϕ[O ∩ U ] → R belongs to S ϕ(p) . So, taking fˆ = f ◦ ϕ−1 in the preceding equation and using (1.2.5), n X ∂ f ◦ ϕ−1 ξi ξˆ f ◦ ϕ−1 = ∂xi i=1

n X → ξ i γ i |p (f ). ϕ(p) = i=1

But recalling how ξˆ was defined, we also have: ξˆ f ◦ ϕ−1 = ξ f ◦ ϕ−1 ◦ ϕ = ξ(f ). Thus, ξ(f ) = n n → → P P ξ i γ i |p . So, as claimed, ξ can be expressed as a linear ξ i γ i |p (f ) for all f in S(p), i.e., ξ =

i=1

i=1

→

combination of the γ i |p .

It follows from proposition 1.2.3, of course, that every vector ξ at p has a unique representation in the n → P ξ i γ i |p . Equivalently, by (1.2.5), form ξ = i=1

ξ(f ) =

n X

→ ξ i γ i |p (f )

i=1

n X

∂ f ◦ ϕ−1 ξ = ∂xi i=1 i

ϕ(p)

(1.2.9)

for all f in S(p). We refer to the coefficients ξ 1 , ..., ξ n as the components of ξ with respect to (U, ϕ). We know that every smooth curve through p determines a vector at p, namely its tangent vector at that point. Using proposition 1.2.3, we can show, conversely, that every vector at p can be realized as the tangent vector of a smooth curve through p. Proposition 1.2.4. Given an n-manifold (M, C), a point p in M , and a vector ξ at p, there is a smooth →

curve γ through p such that γ p = ξ.

13


Proof. Let (U, ϕ) be an n-chart in C with p ∈ U , and let ui (i = 1, ..., n) be the corresponding coordinate maps on U . (Recall that ui = xi ◦ ϕ.) By proposition 1.2.3, we know that there are real numbers ξ 1 , ..., ξ n n → P ξ i γ i |p . Now let γ : I → U be the smooth map defined by: such that ξ = i=1

γ(s) = ϕ−1 u1 (p) + ξ 1 s, ..., un (p) + ξ n s .

Note that γ(0) = p. The exact specification of the domain of γ does not matter, but we may as well take it to be the largest open interval I containing 0 such that, for all s in I, u1 (p) + ξ 1 s, ..., un (p) + ξ n s is in ϕ[U ]. For all f in S(p), ! n X i → ∂ f ◦ ϕ−1 d γ p (f ) = f ◦ γ (0) = ϕ(p) ξ ds ∂xi i=1 =

n X

→

ξ i γ i |p (f ) = ξ(f ).

i=1

→

(The second equality follows by the “chain rule”, and the third by equation (1.2.5).) Thus, γ p = ξ. So far, we have two equivalent characterizations of “vectors” at a point p of a manifold. We can take them to be derivations, i.e., mappings from S(p) to R satisfying conditions (DD1) – (DD3), or take them

to be tangents at p to smooth curves passing through p. We mention, finally, a third characterization that was the standard one before “modern” coordinate-free methods became standard in differential geometry. It requires a bit of preparation. (This third characterization will play no role in what follows, and readers may want to jump to the final paragraph of the section.) Let (U1 , ϕ1 ) and (U2 , ϕ2 ) be n-charts on our background manifold (M, C) such that (U1 ∩ U2 ) 6= ∅.

Let p be a point in (U1 ∩ U2 ). Further, for all i = 1, ..., n, let x′i : ϕ1 [U1 ∩ U2 ] → R be the map defined by x′i = xi ◦ ϕ2 ◦ ϕ1−1 ,

where xi is the ith coordinate map on Rn . We can think of the x′i as providing a second coordinate system on ϕ1 [U1 ∩ U2 ] that is connected to the first by a smooth, invertible transformation (x1 , ..., xn ) 7→ (x′1 (x1 , ..., xn ), ..., x′n (x1 , ..., xn )). Proposition 1.2.5. Under the assumptions of the preceding paragraph, let ξ be a non-zero vector at p whose components with respect to (U1 , ϕ1 ) and (U2 , ϕ2 ) are (ξ 1 , ..., ξ n ) and (ξ ′1 , ..., ξ ′n ). Then the components obey the transformation law ξ ′i =

n X j=1

ξj

∂x′i (ϕ1 (p)). ∂xj

(1.2.10)

(Of course, they also obey its symmetric counterpart, with the roles of xi and ξ i systematically interchanged with those of x′i and ξ ′i .)

14


Proof. Let f be any element of S(p). Then n n −1 −1 X X j ∂ f ◦ ϕ1 ′j ∂ f ◦ ϕ2 ξ ξ ϕ1 (p) = ξ(f ) = ϕ2 (p) . j j ∂x ∂x j=1 j=1

(1.2.11)

Here we have simply expressed the action of ξ on f in terms of the two sets of components (using (1.2.9)). i Hence, in particular, if f = x′i ◦ ϕ1 = xi ◦ ϕ2 ◦ ϕ−1 1 ◦ ϕ1 = x ◦ ϕ2 , we get n X j=1

n

ξj

X ∂xi ∂x′i ξ ′j (ϕ1 (p)) = (ϕ2 (p)) = ξ ′i . j j ∂x ∂x j=1

In what follows, let C(p) be the set of charts in C whose domains contain p. Problem 1.2.1. Let ξ be a non-zero vector at p, and let (k 1 , ..., k n ) be a non-zero element of Rn . Show there exists an n-chart in C(p) with respect to which ξ has components (k 1 , ..., k n ).

(Hint. Consider any n-chart (U1 , ϕ1 ) in C(p), and let (ξ 1 , ..., ξ n ) be the components of ξ with respect to (U1 , ϕ1 ). Then there is a linear map from Rn to itself that takes (ξ 1 , ..., ξ n ) to (k 1 , ..., k n ). Let the n X aij ξ j . Now consider a new chart associated matrix have elements {aij }. So, for all i = 1, ..., n, k i = j=1

(U2 , ϕ2 ) in C(p) where U2 = U1 and ϕ2 is defined by the condition xi ◦ ϕ2 =

n X j=1

aij (xj ◦ ϕ1 ).

Show that the components of ξ with respect to (U2 , ϕ2 ) are (k 1 , ..., k n ).) We have just seen that each vector ξ at p (understood, say, as a derivation) determines a map from C(p) to Rn satisfying the transformation law (1.2.10). (The map assigns to each n-chart the components

of the vector with respect to the n-chart.) It turns out, conversely, that every map from C(p) to Rn

satisfying (1.2.10) determines a unique vector ξ at p. It does so as follows. Let (U1 , ϕ1 ) be an n-chart in C(p). We stipulate that, for all maps f in S(p), n X

∂ f ◦ ϕ−1 1 ξ ξ(f ) = j ∂x j=1 j

ϕ1 (p) ,

(1.2.12)

where (ξ 1 , ..., ξ n ) is the element of Rn associated with (U1 , ϕ1 ). We need only verify that this definition is independent of our choice of n-chart. Let (U2 , ϕ2 ) be any other n-chart in C(p) with associated n-tuple (ξ ′1 , ..., ξ ′n ). Then, by assumption,

the latter are related to (ξ 1 , ..., ξ n ) by (1.2.10). Now consider the map f ◦ ϕ−1 1 : ϕ1 [U1 ∩ U2 ] → R. It can

−1 −1 be realized as the composition of two maps f ◦ ϕ−1 1 = (f ◦ ϕ2 ) ◦ (ϕ2 ◦ ϕ1 ). Hence, by the chain rule, n X ∂ xk ◦ ϕ2 ◦ ϕ−1 ∂ f ◦ ϕ−1 ∂ f ◦ ϕ−1 2 1 1 ϕ1 (p) = ϕ2 (p) ϕ1 (p) j k j ∂x ∂x ∂x k=1 n −1 X ∂ f ◦ ϕ2 ∂x′k ϕ2 (p) ϕ1 (p) , = k j ∂x ∂x k=1

15


for all j. Hence, by (1.2.12) and (1.2.10), " n # n ′k X ∂ f ◦ ϕ−1 X ∂x 2 ξj ξ(f ) = ϕ2 (p) ϕ1 (p) k ∂x ∂xj j=1 k=1   n n n ′k X X X ∂ f ◦ ϕ−1 ∂ f ◦ ϕ2−1 ∂x 2 ′k j  ϕ1 (p)  ϕ2 (p) = ξ ϕ2 (p) . = ξ j k k ∂x ∂x ∂x j=1 k=1

k=1

Thus, our definition of ξ is, indeed, independent of our choice of n-chart. We could equally well have formulated (1.2.12) using (U2 , ϕ2 ) and (ξ ′1 , ..., ξ ′n ). The upshot is that there is a canonical one-to-one correspondence between vectors at p and maps from

C(p) to Rn satisfying (1.2.10). This gives us our promised third (classical) characterization of the former. There is a helpful picture that accompanies our formal account of tangent vectors and tangent spaces. Think about the special case of a 2-manifold (M, C) that is a smooth surface in three-dimensional Eu-

clidean space. In this case, the tangent space to the manifold Mp at a point p is (or can be canonically identified with) the plane that is tangent to the surface at p. In traditional presentations of differential geometry, vectors at points of manifolds are sometimes called “infinitesimal displacements”. The picture suggests where this term comes from. A displacement from p on the surface M is approximated by a tangent vector in Mp . The degree of approximation increases as the displacement on M shrinks. In the limit of “infinitesimal displacements”, the two coincide. (Quite generally, statements about “infinitesimal objects” can be read as statements about corresponding objects in tangent spaces.)

16


1.3

Vector Fields, Integral Curves, and Flows

In what follows, let (M, C) be an n-manifold. (We shall often supress explicit reference to C.) A vector

field on M is a map ξ that assigns to every point p in M a vector ξ(p) in Mp . (Sometimes we shall write ξ|p for the value of the field ξ at p rather than ξ(p).) We can picture it as field of arrows on M . Given any smooth map f : M → R, ξ induces a map ξ(f ) : M → R defined by ξ(f )(p) = ξ|p (f ). If ξ(f ) is smooth for all such f , we say that the vector field ξ itself is smooth.

The proposed picture of a vector field as a field of arrows on M suggests that it should be possible to “thread” the arrows — at least when the field is smooth — to form a network of curves covering M . (See figure 1.3.1.) In fact, this is possible.

M

Figure 1.3.1: Integral curves “threading” the vectors of a smooth vector field.

Let ξ be a smooth vector field on M . We say that a smooth curve γ : I → M is an integral curve of → ξ if, for all s ∈ I, γ γ(s) = ξ γ(s) , i.e., if the tangent vector to γ at γ(s) is equal to the vector assigned by ξ to that point. Intuitively, an integral curve of ξ threads the arrows of ξ and is so parametrized that

it “moves quickly” (i.e., covers a lot of M with each unit increment of the parameter s) where ξ is large, and “slowly” where ξ is small. Let us also say that a smooth curve γ : I → M has initial value p if 0 ∈ I and γ(0) = p. The following is the basic existence and uniqueness theorem for integral curves.

Proposition 1.3.1. Let ξ be a smooth vector field on M and let p be a point in M . Then there is an integral curve γ : I → M of ξ with initial value p that has the following maximality property: if

γ ′ : I ′ → M is also an integral curve of ξ with initial value p, then I ′ ⊆ I and γ ′ (s) = γ(s) for all

s ∈ I ′.

It is clear that the curve whose existence is guaranteed by the proposition is unique. (For if γ ′ : I ′ → M

is another, we have I ′ ⊆ I and I ⊆ I ′ , so I ′ = I, and also γ ′ (s) = γ(s) for all s ∈ I ′ .) It is called the

maximal integral curve of ξ with initial value p. It also clearly follows from the proposition that if γ is an integral curve of ξ with initial value p, and if its domain is R, then γ is maximal. (The converse is false. Maximal integral curves need not have domain R. We shall soon have an example.) The proof of the proposition, which we skip, makes use of the basic existence and uniqueness theorem for solutions to

17


ordinary differential equations. Indeed, the proposition can be understood as nothing but a geometric formulation of that theorem. (See, for example, Spivak [57, volume 1, chaper 5].) Here are some examples. In the following, let x1 and x2 be the standard coordinate maps on R2 . (So if p = (p1 , p2 ) ∈ R2 , then x1 (p) = p1 and x2 (p) = p2 .) ∂ (1) Let ξ be the “horizontal” vector field on R2 . ∂x1 ∂ the vector |p at p assigns to f the number ∂x1 initial value p = (p1 , p2 ) is the map γ : R → R2

Given any point p and any function f in S(p), ∂f (p). The maximal integral curve of ξ with 1 ∂x with

γ(s) = p1 + s, p2 .

∂ is defined similarly.) ∂x2 ∂ ∂ (2) Let ξ be the “rotational” vector field −x2 1 + x1 2 on R2 . The maximal integral curve of ξ ∂x ∂x with initial value p = (p1 , p2 ) is the map γ : R → R2 with (The “vertical” vector field

γ(s) = p1 cos s − p2 sin s, p1 sin s + p2 cos s .

The image of γ is a circle, centered at (0, 0) that passes through p. (In the degenerate case where p is (0, 0), γ is the constant curve that sits at (0, 0).) ∂ ∂ + x2 2 on R2 . The maximal integral curve 1 ∂x ∂x of ξ with initial value p = (p1 , p2 ) is the map γ : R → R2 with

(3) Let ξ be the “radial expansion” vector field x1

γ(s) = (p1 es , p2 es ). If (p1 , p2 ) 6= (0, 0), the image of γ is a radial line starting from, but not containing, (0, 0). If p is

(0, 0), γ is the constant curve that sits at (0, 0).

Let us check one of these — say (2). The indicated curve is, in fact, an integral curve of the given vector field since, for all s ∈ R, and all f ∈ S(γ(s)), by the chain rule, → γ γ(s) (f )

= = = =

d d f ◦ γ (s) = f p1 cos s − p2 sin s, p1 sin s + p2 cos s ds ds ∂f ∂f (γ(s)) −p1 sin s − p2 cos s + 2 (γ(s)) p1 cos s − p2 sin s ∂x1 ∂x ∂f ∂f (γ(s)) −x2 (γ(s)) + 2 (γ(s)) x1 (γ(s)) 1 ∂x ∂x ∂ ∂ −x2 1 + x1 2 (f ). ∂x ∂x |γ(s)

∂ ∂ − x2 2 on R2 . Show that the maximal integral curve 1 ∂x ∂x of ξ with initial value p = (p1 , p2 ) is the map γ : R → R2 with γ(s) = (p1 es , p2 e−s ). (The image of γ Problem 1.3.1. Let ξ be the vector field x1

is a (possibly degenerate) hyperbola satisfying the coordinate condition x1 x2 = p1 p2 .)

18


Next we consider reparametrizations of integral curves. Proposition 1.3.2. Let ξ be a smooth vector field on M , let γ : I → M be an integral curve of ξ, and

let α : I ′ → I be a diffeomorphism taking the interval I ′ to the interval I. Consider the reparametrized curve γ ′ = γ ◦ α : I ′ → M .

(1) If there is a number c such that α(s) = s + c for all s ∈ I ′ , then γ ′ is an integral curve of ξ. (2) Conversely, if γ ′ is an integral curve of ξ and if ξ is everywhere non-zero on γ[I], then there is a number c such that α(s) = s + c for all s ∈ I ′ . Proof. Set t = α(s). For all s ∈ I ′ and all functions f ∈ S(γ ′ (s)), it follows by the chain rule (and the

definition of tangents to curves) that →′ γ γ ′ (s) (f )

= =

That is, for all s ∈ I ′ ,

d d f ◦ γ ′ (s) = f ◦ γ ◦ α (s) = ds ds → dα γ γ(α(s))(f ) (s). ds →′ γ γ ′ (s)

→

= γ γ(α(s))

dα d f ◦γ α(s) (s) dt ds

dα (s). ds

(1.3.1)

Since γ is an integral curve of ξ, we also have → γ γ(α(s))

= ξ γ(α(s))

(1.3.2)

→ for all s ∈ I ′ . Now γ ′ is an integral curve of ξ iff γ ′γ ′ (s) = ξ γ ′ (s) = ξ γ(α(s)) for all s ∈ I ′ . So,

by (1.3.1) and (1.3.2), γ ′ is an integral curve of ξ iff

dα (s) = ξ γ(α(s)) ξ γ(α(s)) ds

(1.3.3)

for all s ∈ I ′ . This equation is the heart of the matter. If there is a c such that α(s) = s + c for all s ∈ I ′ , dα then = 1 everywhere, and so it follows immediately that (1.3.3) holds for all s ∈ I ′ . This gives us ds dα = 1 everywhere. clause (1). Conversely, if (1.3.3) does hold for all s ∈ I ′ , it must be the case that ds (Here we use our assumption that ξ γ(α(s)) is non-zero for all s ∈ I ′ .) So, clearly, α must be of the form α(s) = s + c for some number c. This gives us (2).

The qualification in the the second clause of the proposition — that ξ be non-zero on the image of γ — is necessary. (See problem 1.3.3.) The first clause of the proposition tells us that if γ : I → M is an

integral curve of ξ, then so is the curve defined by setting γ ′ (s) = γ(s + c). We say that γ ′ is derived from γ by “shifting its initial value”. Several useful facts about integral curves follow from proposition 1.3.2 (together with proposition 1.3.1). We list three as problems. The first is a slightly more general formulation of the existence and uniqueness theorem.


19

Problem 1.3.2. (Generalization of proposition 1.3.1) Again, let ξ be a smooth vector field on M , and let p be a point in M . But now let s0 be any real number (not necessarily 0). Show that there is an integral curve γ : I → M of ξ with γ(s0 ) = p that is maximal in this sense: given any integral curve

γ ′ : I ′ → M of ξ, if γ ′ (s0 ) = p, then I ′ ⊆ I and γ ′ (s) = γ(s) for all s in I ′ .

(Hint: Invoke proposition 1.3.1 and shift initial values.)

Problem 1.3.3. (Integral curves that go nowhere) Let ξ be a smooth vector field on M , and let γ : I → M

be an integral curve of ξ. Suppose that ξ vanishes (i.e., assigns the zero vector) at some point p ∈ γ[I].

Then the following both hold.

(1) γ(s) = p for all s in I, i.e., γ is a constant curve. (2) The reparametrized curve γ ′ = γ ◦ α : I ′ → M is an integral curve of ξ for all diffeomorphisms α : I ′ → I.

(Hint: Think about the constant curve, with domain R, that assigns p to all s.) Problem 1.3.4. (Integral curves cannot cross) Let γ : I → M and γ ′ : I ′ → M be integral curves

of ξ that are maximal (in the sense of problem 1.3.2) and satisfy γ(s0 ) = γ ′ (s′0 ). Then the two curves agree up to a parameter shift: γ(s) = γ ′ s + (s′0 − s0 ) for all s ∈ I. Again, let ξ be a smooth vector field on M . We say that ξ is complete if, for every point p in M , the

maximal integral curve of ξ with initial point p has domain R, i.e., is a curve of the form γ : R → M . For example, let M be the restriction of R2 to the vertical strip (p1 , p2 ) : −1 < p1 < 1 , let ξ be the ∂ restriction of the “horizontal” vector field (discussed above) to M , and let p = (0, 0). The maximal ∂x1 integral curve of ξ with initial value p is the map γ : (−1, 1) → M with γ(s) = (s, 0). So ξ is not

complete. (Intuitively, moving along any maximal integral curve of ξ, in either direction, one “runs out ∂ ∂ is complete on M . And of space” in finite parameter time.) In contrast, the “vertical field” 2 ∂x ∂x1 itself is complete when construed as a field on (all of) R2 . Next, let M be the punctured manifold R2 − {(0, 0)}, and let ξ be the restriction of the radial

vector field (the third in our list of examples) to M . Then ξ is complete. This follows directly from our

determination of the maximal integral curves of ξ. It also follows from the assertion in the next problem. (Intuitively, the vectors of ξ rapidly get small as one approaches the puncture point, and so – moving “backward” along a maximal integral curve of ξ – one cannot reach that point in finite parameter time.) Problem 1.3.5. Let ξ be a smooth vector field on M that is complete. Let p be a point in M . Show that the restriction of ξ to the punctured manifold M − {p} is complete (as a field on M − {p}) iff ξ vanishes

at p.

The maximal integral curves of a smooth vector field suggest the flow lines of a fluid. It turns out to be extremely useful to think of them this way. Let ξ be a smooth vector field on the manifold M . We associate with ξ a set Dξ ⊆ R × M and a “flow map” Γ : Dξ → M as follows. We take Dξ to be the

20


set of all points (s, p) with the property that if γ : I → M is the maximal integral curve of ξ with initial

value p, then s ∈ I; and in this case we set Γ(s, p) = γ(s). (That is, if we start at p, and move s units

of parameter distance along the maximal integral curve with initial value p, we arrive at Γ(s, p).) So, in

particular, (0, p) is in Dξ for all p in M , and Γ(0, p) = p for all such. If the vector field ξ is complete, Dξ = R × M . But, in general, Dξ is a proper subset of the latter. (Starting at a point p, it may not

be possible to move s units of parameter distance along the maximal integral curve with initial value p.) We have the following basic result. Proposition 1.3.3. Let ξ be a smooth vector field on M , and let Γ : Dξ → M

be as in the preceding

paragraph. Then Dξ is an open subset of R × M , and Γ is smooth.

The proposition asserts, in effect, that solutions to ordinary differential equations depend smoothly on initial conditions. (See Spivak [57, volume 1, chapter 5].) Assume for the moment that our smooth vector field ξ on M is complete. (So Dξ = R × M .) In this

case, given any s ∈ R, we can define a map Γs : M → M by setting Γs (p) = Γ(s, p). It follows from

proposition 1.3.3 that Γs is smooth. (Γs can be realized as a composite map M → R × M → M with

action p 7→ (s, p) 7→ Γ(s, p) and each of the component maps is smooth.) Furthermore, the indexed set {Γs }s∈R has a natural group structure under the operation of composition Γs ◦ Γt = Γs+t with the identity map Γ0 playing the role of the unit element. (See the next paragraph.) It follows that Γs is

injective and that its inverse (Γs )−1 = Γ−s is smooth. So each Γs is a diffeomorphism that maps M onto itself. We say that {Γs }s∈R is a one-parameter group of diffeomorphisms of M generated by ξ.

Note that, for all p in M , the map from R to M defined by s 7→ Γs (p) is just the maximal integral curve

of ξ with initial point p.

That Γs ◦ Γt = Γs+t for all s and t follows as a consequence of the assertion in problem 1.3.4.

Given any point p in M , and any t ∈ R, let γ : I → M be the maximal integral curve of ξ with Let γ ′ : I ′ → M be the maximal integral initial value Γt (p). Then γ(s) = Γs Γt (p) for all s. curve of ξ with initial value p. Then γ ′ (t) = Γt (p) = γ(0) and γ ′ (s + t) = Γs+t (p) for all s.

Since

γ(0) = γ ′ (t), it follows from the assertion in the problem that γ(s) = γ ′ (s + t) for all s. So we have Γs Γt (p) = γ(s) = γ ′ (s + t) = Γs+t (p) for all p, t, and s.

Now recall the three complete vector fields on R2 considered above. Each defines a one-parameter

group of diffeomorphisms {Γ}s∈R on R2 . The pattern of association is as follows. Field ∂ ∂x1

Associated Diffeomorphisms Γs (p1 , p2 ) = (p1 + s, p2 )

−x2

∂ ∂ + x1 2 1 ∂x ∂x

Γs (p1 , p2 ) = p1 cos s − p2 sin s, p1 sin s + p2 cos s

x1

∂ ∂ + x2 2 ∂x1 ∂x

Γs (p1 , p2 ) = (p1 es , p2 es )


21

In the three cases, respectively, Γs is: a displacement by the amount s in the x1 direction, a (counter clockwise) rotation through s radians with center point (0, 0), and a radial expansion by the factor es with center point (0, 0). Let us now drop the assumption that ξ is complete. Then the “flow maps” Γs : M → M

will not,

in general, be defined for all s. But by paying attention to domains of definition, we can still associate with ξ a set of “local flow maps”. It follows from proposition 1.3.3 that given any point p in M , there is an open interval I ⊆ R containing 0 and an open subset U ⊆ M containing p such that I × U ⊆ Dξ .

If we set Γs (q) = Γ(s, q) for all (s, q) ∈ I × U , then the following all hold. (1) Γs : U → Γs [U ] is a diffeomorphism for all s ∈ I.

(2) (Γs ◦ Γt )(q) = Γs+t (q) for all s, t, and q such that {s, t, s + t} ⊆ I and {q, Γt (q)} ⊆ U . (3) For all q in U , the map γ : I → M defined by γ(s) = Γs (q) is a smooth integral curve of ξ with initial value q.

In this case we say that the collection {Γs : U → Γs [U ]}s∈I is a local one-parameter group of diffeomor-

phisms generated by ξ.

22


1.4

Tensors and Tensor Fields on Manifolds

We start with some linear algebra. We shall return to manifolds shortly. Let V be an n-dimensional vector space. (Here and throughout these notes, “vector spaces” should be understood to be vector spaces over R.) Linear functionals (or covariant vectors or co-vectors) over V are linear maps from V to R. The set of all linear functionals on V has a natural vector space structure. Given two linear functionals α and β, and a real number k, we take α + β and k α to be the linear functionals defined by setting: (α + β)(ξ)

=

α(ξ) + β(ξ),

(k α)(ξ)

=

k α(ξ),

for all ξ in V . The vector space V ∗ of linear functionals on V is called the dual space of V . It is easy 1

2

n

1

2

n

to check that V ∗ has dimension n. If ξ, ξ, ..., ξ form a basis for V , then the elements α, α, ..., α in V ∗ defined by

form a basis for V ∗

j i α ξ = δij 1 2 n called the dual basis of ξ, ξ, ..., ξ.

The vector space V ∗ has its own dual space V ∗∗ consisting of linear maps from V ∗ to R. V ∗∗ is naturally isomorphic to V under the mapping ϕ : V → V ∗∗ defined by setting ϕ(ξ)(α) = α(ξ) for all ξ

in V and all α in V ∗ , i.e., we require that ϕ(ξ) make the same assignment to α that α itself makes to ξ. In our development of tensor algebra we shall use the “abstract index notation” introduced by Roger Penrose. (See Penrose and Rindler [51] for a more complete and systematic treatment.) We start by considering an infinite sequence of vector spaces V a , V b , ..., V a1 , V b1 , ... , all isomorphic to our original n-dimensional vector space V . Here a, b, ..., a1 , b1 , ... are elements of some (unspecified) infinite labeling set and are called “abstract indices”. They must be distinguished from more familiar “counting indices”. We think of isomorphisms being fixed once and for all, and regard ξ a , ξ b , ... as the respective images in V a , V b , ... of ξ in V . The spaces V a , V b , ... have their respective dual spaces (V a )∗ , (V b )∗ , ... . We designate these with lowered indices: Va , Vb , ... .

Our fixed isomorphisms between V and V a , V b , ...

naturally extend to isomophisms between V ∗ and Va , Vb , ... . Given α in V ∗ we take its image in Va to be the unique element αa satisfying the condition αa (ξ a ) = α(ξ) for all ξ in V . It is convenient to drop parentheses and write αa (ξ a ) as αa ξ a or ξ a αa . Thus we have αa ξ a = ξ a αa = αb ξ b = ξ b αb , and so forth. (In what follows, our notation will be uniformly commutative. In a sense, the notation incorporates the canonical isomorphism of V with V ∗∗ . Rather than thinking of ξ a αa as αa (ξ a ), we can think of it as the “action of ξ a on αa ” and understand that as the action on αa of the vector in (V a )∗∗ canonically isomorphic to ξ a .) Indices tell us where vectors and linear functionals reside. So rather than writing, for example, “for all vectors ξ a in V a ...”, it will suffice to write “for all vectors ξ a ...”.

23


We have introduced vector spaces V a , V b , ..., Va , Vb , ... . indexed spaces

...ar V ba11...b s

Now we jump to a larger collection of

(r, s ≥ 1) where the indices a1 , ..., ar , b1 , ..., bs are all distinct. (The order

of superscript indices here will make no difference; nor will that of subscript indices. So, for example,

ad da ad da V bc = V bc = V cb = V cb . But it will make a difference whether particular indices appear in superscript

or subscript position, e.g., V ba 6= V ab .) To keep the notation under control, we shall work first with a

representative special case: V cab .

The elements of this space are multilinear maps that assign real numbers to unordered triples of the form {µa , νb , γ c }, i.e., triples containing one element each from Va , Vb , and V c . (We shall write these

triples, indifferently, as µa νb γ c or νb µa γ c or γ c νb µa or νb γ c µa , and so forth.) By “multilinearity”

we mean that if λ is in V cab , then λ (µa + k ρa )νb γ c λ µa (νb + k τb )γ c

=

λ(µa νb γ c ) + k λ(ρa νb γ c ),

=

λ(µa νb γ c ) + k λ(µa τb γ c ),

λ µa νb (γ c + k δ c ) =

λ(µa νb γ c ) + k λ(µa νb δ c ),

for all µa , ρa , νb , τb , γ c , δ c and all real numbers k. The set V cab has a natural vector space structure. If λ and λ′ are two elements of V cab and k is a real number, we can define new elements (λ + λ′ ) and (k λ) in V cab by setting (λ + λ′ )(µa νb γ c ) (k λ)(µa νb γ c )

= λ(µa νb γ c ) + λ′ (µa νb γ c ), = k λ(µa νb γ c ),

for all µa , νb , γ c . The vector space V cab has dimension n3 . To see this, first note that any triple of vectors {ϕa , ψ b , χc } determines an element in V cab under the rule of association {ϕa , ψ b , χc } : µa νb γ c 7−→ (ϕa µa )(ψ b νb )(χc γ c ). We write this element as ϕa ψ b χc or χc ϕa ψ b or ψ b χc ϕa , and so on. The order of the terms makes 1

2

n

1

2

n

no difference. Next, let ξ a , ξ a , ..., ξ a be a basis for V a with dual basis αa , αa , ..., αa . (Here we have abstract and counting indices side by side.) One can easily verify that the set of all triples of the form i

j

ξ a ξ b αc , with i, j, k ranging from 1 to n, forms a basis for V cab . Thus, every element of V cab can be k

uniquely expressed in the form

n X n n X X

j ijk ia b k

c ξ ξ αc .

i=1 j=1 k=1

Sometimes it will be convenient to recast sums such as this in terms of a single summation index and absorb coefficients, i.e., in the form 3

n X

µa ν b τc . i

i

i

i=1

(Rather than three indices that range from 1 to n, we have one index that ranges from 1 to n3 .)

24


...ar Generalizing now, the tensor space V ba11...b (r, s ≥ 1) consists of multilinear maps assigning real s

numbers to unordered (r + s)-tuples, containing one element each from Va1 , ..., Var , V b1 , ..., V bs . It is a vector space with dimension n(r+s) , and its elements can be realized as linear combinations of the form (r+s) nX

i=1

i

µa1 · · · ν ar γ b1 · · · λbs . i

i

i

We have assumed (r, s ≥ 1). But the definition scheme we have given makes sense, too, when r = 0 and

s = 1, and when r = 1 and s = 0. In the former case, we recover indexed dual spaces as previously

characterized. (The elements of Vb , recall, are just linear maps from V b to R). And in the latter case, we recover our initial indexed vector spaces, at least if we allow for the identification of those spaces with their “double duals”. We can even allow r = s = 0 and construe the tensor space over V with no indices as just R. The elements of tensor spaces are called tensors. Tensor indices in superscript (respectively subscript) position are sometimes called “contravariant” (respectively “covariant”) indices. We have noted that abstract indices give information about where vectors and co-vectors reside, e.g., a

µ belongs to the space V a and νb belongs to Vb . We can extend this pattern of “residence labeling” to elements of arbitrary tensor spaces. For example, we can attach the index configuration of

V cab

and make statements of the form “for all

λab c

ab c

to elements

...”. But things are a bit delicate in the case where

the total number of indices present is greater than one. Though the order of superscript indices and the order of subscript indices make no difference when it comes to labeling tensor spaces, they do make a difference when it comes to labeling tensors themselves. For example, though Vab = Vba , for an arbitrary element αab of that space it need not be the case that αab = αba . (The latter equality captures the condition, not true in general, that the tensor αab is 1

2

n

1

2

n

“symmetric”.) To see why, suppose, once again, that ξ a , ξ a , ..., ξ a is a basis for V a and αa , αa , ..., αa i

j

is its dual basis. Let αab be the element αa αb , for some particular i and j. Then, according to the i

j

Penrose notation (as will be explained), αba is the element αb αa . It follows from what has been said i

j

i

j

so far that the tensors αa αb and αb αa are simply not equal unless i = j. (Why? Assume they are equal. Then they have the same action on all pairs µa ν b . So, in particular, they have the same action i

j

on ξ a ξ b . But

and

So δij = 1, i.e., i = j.)

j j i j i j i i αa αb ξ a ξ b = αa ξ a αb ξ b = 1 j j i i j j i i αb αa ξ a ξ b = αa ξ a αb ξ b = (δij )2 .

A second point about the delicacy of the index notation should be mentioned, though it will not concern us until we reach section 1.9 and work with tensors in the presence of a (non-degenerate) metric gab . We will then want to follow standard practice and use the metric and its inverse g ab to “lower and raise indices”. (The rest of this paragraph can be skipped. It is included only for readers who already

25


know about lowering and raising indices and who may anticipate the problem mentioned here.) For example, we shall write αa b as an abbreviation for αan g nb . A problem will arise, though, when we try to lower or raise an index on a tensor that has indices in both subscript and superscript position. For cd example, do we write λab as λdab or as λadb or as λabd ? The latter three will not, in general, c g

be equal (for the reasons given in the preceding paragraph). To cope with the problem, when the time comes, we shall adopt the convention that superscript indices should never be aligned with subscript indices. Instead, each index will have its own vertical “slot”. So, for example, the elements of the space V cab will carry the index structure

ab

c

or

a b c

or

c

ab

(or

ba

c

or

b a c

or

c

ba

), and we will not assume,

for example, that λab c = λa c b . (For the rest of this section — indeed until section 1.9 — we shall not bother with index slots.) One final preliminary remark about notation is called for. As mentioned before, we want the notation to be uniformly commutative, at least as regards the order of tensors within an expression (in contrast c ab to the order of indices within a tensor). So, for example, the number λab c (µa νb γ ) that the tensor λc c or as νb γ c λab or as µa νb γ c λab assigns to a triple µa νb γ c will be written as λab c µa , and c c µa νb γ c a b c a b ab so forth. Furthermore, if λab c is the tensor ϕ ψ χc , we shall write λc (µa νb γ ) as ϕ ψ χc µa νb γ

or as χc µa ψ b νb ϕa γ c or as any other string with the individual vectors in some order or other. The order does not matter because it is the indices here that determine the crucial groupings: ϕa with µa , ψ b with νb , χc with γ c . ...ar We now have in hand the various tensor spaces V ba11...b . Within each one (just because it is a vector s

space), there is an addition operation that is associative and commutative. We will be interested in three other tensor operations: exterior multiplication, index substitution, and contraction. We will consider them in turn. “Exterior multiplication” (or, perhaps, “tensor multiplication”), first, is an operation of structure ...ar a1 ...ar c1 ...cm m V ba11...b × V dc11...c ...dn → V b1 ...bs d1 ...dn , s

where the indices a1 , ..., ar , b1 ...bs , c1 , ..., cm , d1 , ..., dn are all distinct. It is defined in an obvious way. Consider a representative special case ab V cab × Vf d → V cf d. ab ab The exterior product of αab c and ξf d , written αc ξf d or ξf d αc , is defined by setting

c c f d ξf d µf ν d = αab αab c λa ρb δ c ξf d λa ρb δ µ ν

for all λa , ρb , δ c , µf , ν d . As usual, generally we shall drop parentheses and write terms in any order. c f d c f d So the action of αab will be expressed, indifferently, as αab c ξf d on λa ρb δ µ ν c λa ρb δ ξf d µ ν c f d d or as αab or as λa ρb ξf d δ c µf αab c ξf d λa ρb δ µ ν c ν , and so forth. It should be clear that exterior

multiplication, as defined here, is commutative, associative, and distributive over addition. Notice, also, that our notation is consistent. Consider, for example, the expression τ a εb ϕc αa βb γ c . We can construe

26


it as the action of τ a εb ϕc on αa βb γ c , or as the action of εb ϕc αa on τ a βb γ c , or as the action of γ c ϕc τ a εb on αa βb , and so on. (The third reading makes sense: γ c ϕc τ a εb is the element of V ab that arises if one multiples the element τ a εb by the number γ c ϕc .) Each of these functional operations yields the same number, so no consistency problem arises. The operation of “(x → y) index substitution” has the structure ...ar ...ar V bx1 ab21...b → V by1ab21...b s s

...ar ...ar or V xa1ba12...b → V ya1ba12...b , s 2

where the indices x, y, a1 , ..., ar , b1 , ..., bs are all distinct. In defining the operation, it is, again, easiest to consider a representative special case, say V cab → V cdb . Given a tensor αab c , it can be expressed as a

sum of the form

3

αab c

=

n X

µa ν b τc . i

i

i

i=1

db We take the result of (a → d) index substitution on αab c , which we write as αc , to be the sum 3

αdb c =

n X

µd ν b τc . i

i

i

i=1

i

(This makes sense because we already have a fixed isomorphism between V a and V d that takes each µa i

to µ d .) Of course, it must be checked that this definition is independent of the choice of expansion for αab c . That is, one must check that if 3

n X i=1

then

3

n X i=1

i

3

n X i i i δ a εb ρc , µ ν τc = i

a

i

b

i

i=1 3

n X i i i δ d ε b ρc . µ ν τc = i

d

i

i

d

b

i

i=1

a

i

i

But this follows from the fact that µ λa = µ λd and δ a λa = δ d λd for all i and all λa . It can easily be checked that index substitution commutes with addition, exterior multiplication, and ab ab db db db ab ab other index substitutions. For example, if αab c = βc +γc , then αc = βc +γc . If λcf g = αc ξf g , then db λdb cf g = αc ξf g . And the tensor that results from first applying (a → b) index substitution and then

(c → d) index substitution to αacf is the same as that resulting from reversing the order and applying

first (c → d) index substitution and then (a → b) index substitution. It is written as αbdf . All these

facts, in a sense, are built into our notation.

Our final tensor operation, “(x, y) contraction”, has the structure ...ar ...ar V yx ba11...b → V ba11...b , s s

where the indices x, y, a1 , ..., ar , b1 , ..., bs are all distinct. Consider, for example, (a, c) contraction with action V cab → V b . Suppose

3

αab c =

n X i=1

µa ν b τc . i

i

i

27


We take the result of applying (a, c) contraction to αab c to be 3

αab a

=

n X

µa τa ν b . i

i

i

i=1

n P

3

3

3

This could also be expressed as

i

i

n P

i

µc τc ν b or as

i

i

i

i

i

µc ν b τc or as

i=1

i=1

i

i

of the listed possibilities is equal to the first because µ a τa = µ d τd

n P

i

i

i

µd τd ν b , and so forth. The last i=1 i i for all µ a and τa . We write this

cb db result as αab a (or αc or αd , and so forth.). It is important that contracted indices on a tensor, i.e., ones

that appear in both contravariant and covariant position, play no role in determining the space in which b ab the tensor resides. αab a belongs to V , not some space Va . Indeed, there is no such space as we have set

things up. To prove that contraction is well defined, i.e., independent of one’s choice of expansion, a simple lemma is needed. k

k

Lemma 1.4.1. For all r ≥ 1, and all ϕa and ψ c (k = 1, ..., r), r X k k ϕa ψ c = 0

r X k k ϕa ψ a = 0.

=⇒

k=1

k=1 1

2

n

1

2

n

Proof. Let ξ a , ξ a , ..., ξ a be a basis for V a with dual basis αa , αa , ..., αa . Then, for each k = 1, ..., r, n n i k P P j k dkj αc . Assume cki ξ a and ψ c = there exist numbers cki and dkj (i, j = 1, ..., n) where ϕa = j=1

i=1

the left-side condition holds. Then for all l = 1, ..., n, 0

=

r X k k ϕa ψ c

k=1

=

r X n X n X

!

l

αa ξ c = l

r n X X i=1

k=1

cki dkj δil δjl =

k=1 i=1 j=1

r X

 ! n X l j l cki ξ a  dkj αc  αa ξ c i

j=1

ckl dkl .

k=1

It follows that the right side condition holds, since  ! n r X n X n r r n X X X X X i k j k cki dkj δij dkj αa  = cki ξ a  ϕa ψ a = k=1

k=1

=

i=1

r n X X

j=1

k=1 i=1 j=1

cki dki = 0.

i=1 k=1

Each term

r P

k=1

cki dki in the final sum is 0 by the calculation just given.

Problem 1.4.1. Show that lemma 1.4.1 can also be derived as a corollary to the following fact (Herstein [32, p. 272]) about square matrices: if M is an (r × r) matrix (r ≥ 1) and M 2 is the zero matrix, then i

j

the trace of M is 0. (Hint: Consider the r × r matrix M with entries Mij = ϕa ψ a .)

28


k

k

k

Corollary 1.4.2. For all r ≥ 1, and all β a , γ b , ψ c (k = 1, ..., r), r X k k k β a γ b ψc = 0

=⇒

k=1

Proof.

r X k k k β a ψ a γ b = 0.

k=1

It follows from the left-side condition that, for all λb ,

r P

k=1

k k k β a γ b λb ψ c = 0.

Applying the

r k k k P k k k β a ψ a γ b λb = 0. But here λb lemma with ϕa = γ b λb β a for all k = 1, ..., r , we may infer that k=1

is arbitrary. So it must be the case that the right-side condition holds.

It follows immediately that contraction is well defined for our tensor αab c . For if 3

αab c

=

n X i=1

3

n X i i i δ a χb ρ c , µ ν τc = i

a

i

b

i

i=1

3

3

we can apply corollary 1.4.2 to the difference

n P

i=1

i

i

i

µ a ν b τc −

n P

i=1

i

i

i

δ a χ b ρc (construed as a sum over 2n3

terms). And the corollary can be recast easily for tensors with other index structures. The contraction operation commutes with addition, exterior multiplication, index substitution, and other contractions. Note, once again, the consistency of our notation. The expression β a γa , for example, can be construed as the action of the functional γa on β a , or as the exterior product of β a with γb followed by (a, b) contraction, or as the exterior product β b with γa followed by (a, b) contraction, and so forth. There is no need to choose among these different readings. Similarly, αac λa σ c can be understood as the action of αac on λa σ c , or as the exterior product of αac with λb σ d followed by (a, b) and (c, d) contractions, or as the action of λa on αac σ c , and so forth. The operations we have introduced on tensors may seem a bit complex. But one quickly gets used to them and applies them almost automatically where appropriate. That is one of the virtues of the abstract index notation. One gets to manipulate tensors as easily as one manipulates components of tensors in traditional tensor analysis. One has the best of both worlds: complete basis (or coordinate) independence, and the computational convenience that comes with indices. Two bits of special notation will be useful. First, we introduce the “delta tensor” δba . It is the element of V ba defined by setting δba ηa ξ b = ηa ξ a for all ηa and ξ b . (Clearly, δba , so defined, is a tensor since it is linear in both indices.) Notice that the defining condition is equivalent to the requirement that δba ξ b = ξ a for all ξ b , and also to the requirement that δba ηa = ηb for all ηa . We can think of δba as an (a → b)

index substitution operator acting on contravariant indices, or as a (b → a) index substitution operator n3 P i i i ac bc µb ν c τd . acting on covariant indices. So, for example, δba αbc d = αd . To see this, suppose that αd = i=1

Then

3

δba αbc d =

n X i=1

3

n X i i i i i i µa ν c τd = αac δba µb ν c τd = d . i=1

29


1

2

n

1

2

n

Given a basis ξa , ξa , ..., ξa for V a with dual basis αa , αa , ..., αa , δba can be expressed as δba =

n P

i

i

ξa αb .

i=1

(This follows since the left- and right-side tensors in this equation have the same action on the basis 1

2

n

elements ξ b , ξ b , ..., ξ b .) It follows that δaa = n. The second bit of useful notation is for “symmetrization” and “anti-symmetrization” of tensors. Consider, for example, the tensor β ab . Corresponding to it is the tensor β ba . One can think of the latter as arising from the former by a series of index substitutions: β ab → β cb → β cd → β bd → β ba .

have already discussed the fact that, though β

ab

and β

ba

belong to V

ab

(We

, in general it is not the case that

β ab = β ba .) We take β (ab) and β [ab] to be the respective symmetrization and anti-symmetrization of β ab : β (ab)

=

β [ab]

=

1 ab (β + β ba ) 2 1 ab (β − β ba ). 2

b Similarly, given a tensor γcdg , we set b γ(cdg)

=

b γ[cdg]

=

1 b b b b b b (γ + γgcd + γdgc + γcgd + γgdc + γdcg ) 6 cdg 1 b b b b b b (γ + γgcd + γdgc − γcgd − γgdc − γdcg ). 6 cdg

In general, a tensor with round brackets surrounding a collection of p consecutive indices (all contravariant 1 or all covariant) is to be understood as times the sum of the p! tensors obtained by taking the selected p! indices in all possible permutations. (Each permutation can be achieved by multiple index substitutions.) In the case of square brackets, the only difference is that each term in the sum receives a coefficient of (+1) or (−1) depending on whether the indices in that term form a positive or negative permutation of the original sequence. The operations of symmetrization and anti-symmetrization commute with addition, exterior multiplication, and index substitution. So, for example, if β ab = γ ab + ρab , then b b β (ab) = γ (ab) + ρ(ab) . If γcdg = λcdg ξ b , then γ(cdg) = λ(cdg) ξ b . And if one applies (c → f ) index

b substitution to γcdg and then symmetrizes over the indices f, d, and g, the resulting tensor is the same

one obtained if one first symmetrizes over c, d, and g, and then applies (c → f ) index substitution. ...ar We say that a tensor of the form αba11...b is (totally) symmetric in indices b1 , ..., bs if interchanging s ...ar ...ar = αba11...b . any two of these indices leaves the tensor intact, or, equivalently, if α(ba11 ...b s s)

We say it is

(totally) anti-symmetric in those indices if the interchange in each case has the effect of multiplying the ...ar ...ar = αba11...b . (The conditions of symmetry and anti-symmetry tensor by (−1) or, equivalently, if α[ba11...b s s]

in indices a1 , ..., ar are defined similarly.) The following proposition will be useful in what follows. Proposition 1.4.3. If ...ar (1) αba11...b is symmetric in indices b1 , ..., bs , and s ...ar b1 (2) αba11...b ξ ...ξ bs = 0 for all ξ in V , s ...ar ...ar then αba11...b = 0. (A parallel proposition holds if αba11...b is symmetric in indices a1 , ..., ar .) s s

30


Proof. We prove the proposition by induction on s. The case s = 1 is trivial. So assume s > 1 and assume the proposition holds for s − 1. For all vectors µ and ν in V , and all real numbers k, we have, ...ar by (2): 0 = αba11...b (µ + k ν)b1 ... (µ + k ν)bs . Expanding the right side of the equation and using (1), we s

arrive at 0

s ...ar b1 ...ar b1 bs + k αba11...b µ ... µbs−1 ν bs + ... ... µ = αba11...b µ s s 1 s ...ar b1 ...ar b1 b2 + ν ... ν bs . k s−1 αba11...b µ ν ... ν bs + k s αba11...b s s s−1

But k is arbitrary here. The only way the right-side sum can be 0 for all values of k is if each of the ...ar b1 terms in the sum (without the coefficient) is 0. In particular, αba11...b µ ... µbs−1 ν bs = 0. s ...ar ...ar ...ar bs is completely symmetric in the indices α′ ba11...b = αbas1...b ν . The tensor α′ bas1...b s−1 s−1 s a ...a r µb1 ... µbs−1 = 0 for all µ in V . So, by our induction hypothesis, it must α′ b11...bs−1 ...ar a1 ...ar bs αb1 ...bs ν = 0. But ν was an arbitrary vector. So αba11...b = 0, as claimed. s

Now let

b1 , ..., bs−1 , and be the case that

Sometimes it will be convenient to work with this proposition in a slightly more general form. Let Σ and Π be strings of indices, possibly empty, in which a1 , ..., ar , b1 , ..., bs do not appear. Then we ...ar a1 ...ar r can say that a tensor αΣba11 ...b is (totally) symmetric in indices b1 , ..., bs if αΣ(ba11...a ...bs )Π = αΣb1 ...bs Π . sΠ

(The case of (total) anti-symmetry is handled similarly.) It follows as a corollary to the proposition that ...ar ...ar if n-dimensional αΣba11 ...b is symmetric in indices b1 , ..., bs , and if αΣba11 ...b ξ b1 ... ξ bs = 0 for all ξ sΠ sΠ ...ar in V , then αΣba11 ...b = 0. sΠ

(It follows because we can always contract on all the indices in Σ and Π

with arbitrary, distinct vectors and generate a tensor to which the proposition is directly applicable.) Of course, a similar generalization of the proposition is available in the case where the “extra indices” are in covariant position. This completes our discussion of tensor algebra. We now return to manifolds. Suppose (M, C) is an

n-manifold and p is a point in M . Then Mp is an n-dimensional vector space. We can take it to be our

fundamental space V and construct a hierarchy of tensor spaces over it. A tensor field on M is simply an assignment of a tensor (over Mp ) to each point p in M , where the tensors all have the same index structure. So, for example, a vector field ξ a on M (as defined in section 1.3) qualifies as a tensor field on M . The tensor operations (addition, exterior multiplication, index substitution, and contraction) are all applied pointwise, and so they extend naturally to tensor fields. We already know what it means for a scalar field or a (contravariant) vector field on M to be smooth. We now take a covariant vector field αa on M to be smooth if (ξ a αa ) is smooth for all smooth vector fields ...ar ...ar b1 ξ a on M . Quite generally, we say that a tensor field λba11...b on M is smooth if λba11...b ξ ... η bs αa1 ... βar s s

is smooth for all smooth fields ξ b1 , ..., η bs , αa1 , ..., βar on M . This pattern of definition is extremely common. One starts with a concept (in this case smoothness) applicable to scalar fields, then extends it to contravariant vector fields by considering their action on scalar fields, then extends it to covariant vector fields by considering their action on contravariant fields,


31

then extends it to tensor fields of arbitrary index structure by considering their action on (appropriate combinations of) contravariant and covariant vector fields. It follows from the definition of smoothness for tensor fields just given that the four tensor operations take smooth tensor fields to smooth tensor fields.

32


1.5

The Action of Smooth Maps on Tensor Fields

In this section, we consider when and how it is possible to use a smooth map between manifolds to carry tensors at a point, and tensor fields, from one manifold to the other. We start with tensors at a point. Let (M, C) and (M ′ , C ′ ) be manifolds, not necessarily of the same

dimension, let ψ : M → M ′ be a smooth map of M into M ′ , and let p be a point in M . There is no

natural way to transfer arbitrary tensors between p and ψ(p) — at least not without further assumptions in place. But it is possible to associate with ψ two restricted transfer maps. Let us say that a tensor (at some point on some manifold) is contravariant (resp. covariant ) if all of its indices are in contravariant (resp. covariant) position. The rank of such a tensor is the number of its indices. We allow the number to be 0, i.e., we regard scalars (real numbers) as both contravariant and covariant tensors of rank 0. The first of our two restricted transfer maps, the “push-forward map” (ψp )∗ , takes contravariant tensors at p to contravariant tensors of the same rank at ψ(p). The second, the “pull-back map” (ψp )∗ , takes covariant tensors at ψ(p) to covariant tensors of the same rank at p. We define (ψp )∗ and (ψp )∗ in four stages. (For clarity, we mark objects defined on M ′ with a prime.) (Stage 0) Given any real number c, we set (ψp )∗ (c) = (ψp )∗ (c) = c. (Stage 1) Given a vector ξ a at p, we define (ψp )∗ (ξ a ) at ψ(p) as follows. Let α′ : O′ → R be an element of S ψ(p) . Then (α′ ◦ ψ) : ψ −1 [O′ ] → R is an element of S(p). We need to specify what assignment (ψp )∗ (ξ a ) makes to α′ . We set

(ψp )∗ (ξ a ) (α′ ) = ξ a (α′ ◦ ψ).

(1.5.1)

This makes sense because (α′ ◦ ψ) is an object of the sort to which ξ a makes assignments. (Stage 2) Next, consider a covariant tensor ηb′ 1 ... bs at ψ(p). We define the pull-back tensor (ψp )∗ (ηb′ 1 ... bs ) 1

s

at p by specifying its action on arbitrary vectors ξ b1 , ..., ξ bs there. We set

1 s s 1 (ψp )∗ (ηb′ 1 ... bs ) ξ b1 ... ξ bs = ηb′ 1 ... bs (ψp )∗ (ξ b1 ) ... (ψp )∗ (ξ bs ) .

(1.5.2)

Here, of course, we understand the right side because we know (from stage 1) how to push forward the i

vectors ξ bi . (Stage 3) Finally, consider a contravariant tensor ξ a1 ... ar at p with r ≥ 2. We define the push-forward 1

r

tensor (ψp )∗ (ξ a1 ... ar ) at ψ(p) by specifying its action on arbitrary vectors η ′a1 , ..., η ′ar there:

1 r 1 r (ψp )∗ (ξ a1 ... ar ) η ′a1 ... η ′ar = ξ a1 ... ar (ψp )∗ (η ′a1 ) ... (ψp )∗ (η ′ar ) .

This completes the definition of (ψp )∗ and (ψp )∗ .

Several basic facts about them are recorded in the next proposition.

(1.5.3)

33


Proposition 1.5.1. Let ψ : M → M ′ be a smooth map of the manifold M into the manifold M ′ . Let

p be any point in M . Then (ψp )∗ and (ψp )∗ have the following properties. (1) (ψp )∗ and (ψp )∗ commute with addition. For example, (ψp )∗ (ξ abc + ρabc ) = (ψp )∗ (ξ abc ) + (ψp )∗ (ρabc ). (2) (ψp )∗ and (ψp )∗ commute with exterior multiplication. ′ ′ For example, (ψp )∗ (ηabc µ′de ) = (ψp )∗ (ηabc ) (ψp )∗ (µ′de ) .

(3) (ψp )∗ and (ψp )∗ commute with index substitution.

(4) For all tensors ξ a1 ...ar c1 ...cs and ρb1 ...br at p, and all tensors ηa′ 1 ...ar and µ′b1 ...br d1 ...ds at ψ(p), (ψp )∗ (ξ a1 ...ar c1 ...cs ) ηa′ 1 ...ar = (ψp )∗ ξ a1 ...ar c1 ...cs ((ψp )∗ (ηa′ 1 ...ar )) , (1.5.4) (1.5.5) (ψp )∗ (µ′b1 ...br d1 ...ds ) ρb1 ...br = (ψp )∗ µ′b1 ...br d1 ...ds ((ψp )∗ (ρb1 ...br )) . Note that we cannot replace clause (4) with the simpler assertion that (ψp )∗ and (ψp )∗ commute with contraction. We cannot claim, for example, that (ψp )∗ (ξ ac ηa ) = (ψp )∗ (ξ ac ) (ψp )∗ (ηa ) , since the second term on the right side is not well formed. The push-forward map (ψp )∗ makes assignments only

to contravariant vectors at p. Note also that it follows as a special case of clause (2) that (ψp )∗ and (ψp )∗ commute with scalar multiplication. For example, (ψp )∗ (c ξ ab ) = ((ψp )∗ (c)) (ψp )∗ (ξ ab ) = c (ψp )∗ (ξ ab ) . So, clearly, (ψp )∗ and (ψp )∗ are linear maps (when restricted to tensors of a fixed rank).

Proof. All four clauses in the proposition follow easily from the definitions of (ψp )∗ and (ψp )∗ . For the fourth clause, one first considers contractions involving (contravariant or covariant) vectors, i.e., ((ψp )∗ (ξ a c1 ...cs )) ηa′ ρ

b1 ...br

or

((ψp )∗ (ηb′ d1 ...ds )) ρb ,

and then uses the fact that every tensor

ηa′ 1 ...ar

or

can be represented as a sum over products of such vectors. The desired conclusion then follows

from clauses (1) and (2). By way of example, let us verify one instance of the fourth clause, say: (ψp )∗ (ξ ac ) ηa′ = (ψp )∗ ξ ac (ψp )∗ (ηa′ ) .

To show that the two (right- and left-side) vectors at ψ(p) are equal, it suffices to demonstrate that they

have the same action on any vector µ′c there. But this follows, since (ψp )∗ ξ ac (ψp )∗ (ηa′ ) µ′c = ξ ac (ψp )∗ (ηa′ ) (ψp )∗ (µ′c ) = (ψp )∗ (ξ ac ) ηa′ µ′c .

Both equalities are instances of (1.5.3). The role of ξ a1 ...ar is played by ξ ac (ψp )∗ (ηa′ ) in the first, and by ξ ac in the second.

Now we turn our attention to fields on M and M ′ . At each point p in M , we have transfer maps (ψp )∗ and (ψp )∗ . The question arises whether they can be “aggregated” to carry contravariant fields on M to ones on M ′ or, alternatively, carry covariant fields on M ′ to ones on M . Here an asymmetry arises.

34


Consider first a tensor field ξ a1 ...ar on M . For all p in M , (ψp )∗ ξ a1 ...ar (p) is a tensor at ψ(p).

(ξ a1 ...ar (p) is the value of the field at p, and it is pushed forward by (ψp )∗ .) But these individual assignments do not, in general, determine a field on M ′ . For one thing, if ψ is not injective, there will be distinct points p and q such that ψ(p) = ψ(q), and nothing guarantees that (ψp )∗ ξ a1 ...ar (p) = (ψq )∗ ξ a1 ...ar (q) . Furthermore, even if ψ is injective, this prescription will not transfer a tensor to a point p′ in M ′ unless it is in the range of ψ, i.e., unless p′ = ψ(p) for some p in M .

But no problems arise if we work in the other direction. Consider a field ηb′ 1 ...bs on M ′ . Then at every point p, there is a well-defined pull-back tensor (ψp )∗ ηb′ 1 ...bs ψ(p) . It just does not matter whether

ψ is injective or whether its range is all of M ′ . So we can aggregate the individual pull-back maps at different points to generate a map ψ ∗ that takes covariant tensor fields on M ′ to ones on M of the same rank. In particular, ψ ∗ takes scalar fields α′ : M ′ → R on M ′ to scalar fields ψ ∗ (α′ ) = (α′ ◦ ψ)

(1.5.6)

on M . (Think about it this way. The pull-back field ψ ∗ (α′ ) assigns to any point p in M the same number that α′ assigns to ψ(p). (Recall the 0-th stage in the definition of (ψp )∗ .) So, for all p in M , ψ ∗ (α′ )(p) = α′ (ψ(p)) = (α′ ◦ ψ)(p).) Three of the (pointwise) algebraic conditions listed in proposition 1.5.1 carry over immediately. Thus, ψ ∗ commutes with addition, exterior multiplication, and index substitution (if these are now understood as operations on tensor fields rather than as operations on tensors at a point). The fourth condition, the one involving contraction, does not carry over because it makes reference to individual push-forward maps (ψp )∗ (and these, we know, cannot, in general, be aggregated). In addition, ψ ∗ satisfies a natural smoothness condition; namely, it takes smooth fields on M ′ to smooth fields on M . This is immediate for the case of scalar fields. (If α′ : M ′ → R is smooth, then certainly the composed map ψ ∗ (α′ ) = (α′ ◦ ψ) is smooth as well.) But a short detour will be required for the other cases.

Let us temporarily put aside our map ψ between manifolds and consider a general fact about the representation of covariant vector fields on a manifold M . Given any smooth scalar field α : M → R,

we associate with it a smooth covariant vector field da α on M , called its “exterior derivative”. (Here we partially anticipate our discussion of exterior derivative operators in section 1.7.) It is defined by the

requirement that, for all p in M and all vectors ξ a at p, ξ a da α = ξ(α), i.e., ξ a da α is the directional derivative of α at p in the direction ξ a . (The condition clearly defines a covariant vector, i.e., a linear functional over Mp , at each point p. And the resultant field da α is smooth since, given any smooth vector field ξ a on M , ξ a da α is a smooth scalar field on M .) The fact we need is the following. Lemma 1.5.2. Let λa be a smooth field on an n-dimensional manifold (M, C). Then, given any point p 1

n 1

n

in M , there exists an open set O containing p, and smooth real-valued maps f , ..., f , g, ..., g on O, such 1

1

n

n

that λa = f da g + ... + f da g on O.

35


Proof. Let p be a point in M , let (O, ϕ) be a chart in C with p ∈ O, and let u1 , ..., un be the associated →

→

coordinate maps on O. At every point q in O, the coordinate curve tangent vectors ( γ 1|q )a , ..., ( γ n|q )a

associated with u1 , ..., un form a basis for (Mq )a . (Recall proposition 1.2.3.) Now consider the vector →

fields da u1 , ..., da un on O. We claim that they determine a dual basis at every q, i.e., ( γ i|q )a (da uj ) = δij →

→

for all i, j ∈ {1, ..., n}. Indeed, this follows immediately since ( γ i|q )a (da uj ) = γ i|q (uj ) (by the definition of 1

→

n

da ) and γ i|q (uj ) = δij (by equation (1.2.6)). So we can express λa in the form λa = f da u1 + ... + f da un i

1

→

n

on O, where f = ( γ i )a λa . The coordinate maps u1 , ..., un are certainly smooth. And the maps f , ..., f →

→

must be smooth as well since λa and the coordinate tangent fields ( γ 1 )a , ..., ( γ n )a are so. With the lemma in hand, let us return to the original discussion. Again, let ψ be a smooth map from the manifold M into the manifold M ′ . Note that given any smooth field α′ : M ′ → R on M ′ , we have ψ ∗ (da α′ ) = da ψ ∗ (α′ ) .

(1.5.7)

(To see this, let p be any point in M and let ξ a be any vector at p. Then ξ b ψ ∗ (db α′ ) |p = (ψp )∗ (ξ b ) (db α′ )|ψ(p) = (ψp )∗ (ξ b ) (α′ ) = ξ b (α′ ◦ ψ) = ξ b db (ψ ∗ (α′ )) |p .

The first equality is an instance of (1.5.2), with (db α′ )|ψ(p) playing the role of ηb′ ; the third is an instance of (1.5.1). The second follows from the definition of the operator da , and the fourth from that definition together with (1.5.6). So (1.5.7) holds at all points p in M .) It is our goal, once again, to show that, for all smooth fields ηb′ 1 ...bs on M ′ , the pull-back field ψ ∗ (ηb′ 1 ...bs ) on M is smooth as well. Consider the case of a smooth vector field ηb′ on M ′ . Suppose M ′ has dimension n. We know from the lemma that given any point p′ in M ′ , we can find an open set O′ containing p′ in n i X i f ′ db g ′ (with the constituent maps all smooth). Hence, we which ηb′ admits the representation ηb′ = i=1

have

ψ ∗ (ηb′ ) = ψ ∗

n X i=1

throughout ψ

−1

′

i

i

f ′ db g ′

!

=

n X

i

i

ψ ∗ (f ′ ) ψ ∗ (db g ′ ) =

n X i=1

i=1

i i ψ ∗ (f ′ ) db ψ ∗ (g ′ )

[O ]. (We get the second equality from the fact that ψ ∗ respects the tensor operations of

addition and exterior multiplication (in the sense discussed above). The third follows from (1.5.7).) But the constituent fields in the far right sum are all smooth. (We have already seen that ψ ∗ takes smooth scalar fields to smooth scalar fields.) So ψ ∗ (ηb′ ) itself is smooth on ψ −1 [O′ ]. But as p′ ranges over M ′ , the corresponding pull-back sets ψ −1 [O′ ] cover M . It follows that ψ ∗ (ηb′ ) is smooth on (all of) M . It remains to consider the general case – smooth fields on M ′ of the form ηb′ 1 ...bs . But this case quickly reduces to the preceding one. We can express any such field, at least locally, in the form s

ηb′ 1 ...bs

=

n X i=1

i

i

µ ′b1 ... ν ′bs ,

36


i

i

where µ ′b1 , ..., ν ′bs (i = 1, ..., ns ) are all smooth fields on M ′ . Since the individual pull-back fields i

i

ψ ∗ (µ ′b1 ), ..., ψ ∗ (ν ′bs ) are smooth (and since ψ ∗ commutes with addition and exterior multiplication), it follows that ψ ∗ (ηb′ 1 ...bs ) must be smooth on M . In summary, we have established the following. Proposition 1.5.3. Let ψ : M → M ′ be a smooth map of the manifold M into the manifold M ′ . Then

ψ ∗ is a map from smooth covariant tensor fields on M ′ to smooth covariant fields on M of the same rank that commutes with addition, exterior multiplication, and index substitution and that also satisfies (1.5.7). The complications and asymmetries we have encountered all have their origin in the fact that we have only been assuming that ψ is a smooth map of M into M ′ . Now, finally, let us consider the case where ψ is, in fact, a diffeomorphism of the first onto the second, i.e., there is a well-defined inverse map ψ −1 : M ′ → M that is also smooth. Then, as one would expect, there is induced a natural one-to-one

correspondence between smooth tensors fields of arbitrary index structure on the two manifolds, and this correspondence fully respects the four tensor operations. We already know how ψ ∗ acts on smooth covariant tensor fields (and scalar fields) on M ′ . Now we can characterize its action on a smooth field ...ar λ′ ba11...b s

of unrestricted index structure on M ′ . We stipulate that, given any point p in M , and any 1

r

1

s

smooth fields η a1 , ..., η ar , ξ b1 , ..., ξ bs on M , s

...ar ...ar ) ψ ∗ (λ′ ba11...b ) (η a1 ... ξ bs )|p = (λ′ ba11...b s |ψ(p) s |p 1

s 1 (ψ)∗ (η a1 ) ... (ψ)∗ (ξ bs )

|ψ(p)

.

(1.5.8)

Of course, the right side makes sense only if we understand how ψ∗ acts on smooth vector fields ηa and ξ b on M . But we do understand (in this new context where ψ is a diffeomorphism). Here we can aggregate the individual push-forward maps (ψp )∗ to generate a map ψ∗ that knows how to act on contravariant vector fields — just as previously we aggregated the maps (ψp )∗ to generate a map ψ ∗ that knows how to act on covariant vector fields. And we can take ψ∗ (ηa ) to be (ψ −1 )∗ (ηa ). This completes the definition of ψ ∗ . Notice that this general characterization of ψ ∗ reduces to the one given previously in the special case where it acts on a covariant field λ′ b1 ...bs . ...ar The way to remember 1.5.8 is this. A trade-off is involved. Pulling back λ′ ba11...b from ψ(p) to p, and s

having it act there on particular vectors yields the same result as pushing those vectors forward from p ...ar to ψ(p) and having λ′ ba11...b act on them there. s

We have just seen how to extend ψ ∗ so that it acts on smooth fields on M ′ of unrestricted index structure (when ψ is a diffeomorphism). Of course, we can extend ψ∗ similarly. Indeed, we can take it to be (ψ −1 )∗ . It is a straightforward matter to confirm that ψ ∗ (and so ψ∗ ) commutes with addition, exterior multiplication, contraction, and index substitution. By way of example, we verify that, for all smooth


37

fields α′ab and ξ ′b on M ′ , ψ ∗ (α′ab ξ ′b ) = ψ ∗ (α′ab ) ψ ∗ (ξ ′b ).

(1.5.9)

Let ηa be any smooth field on M . Then, invoking (1.5.8) and dropping explicit reference to points of evaluation, we have ψ ∗ (α′ab ) ψ ∗ (ξ ′b ) ηa = α′ab ψ∗ ψ ∗ (ξ ′b ) ψ∗ (ηa ) = α′ab ξ ′b ψ∗ (ηa ) = ψ ∗ (α′ab ξ ′b ) ηa .

(For the second equality, we use the fact that (ψ∗ ◦ ψ ∗ ) = the identity map.) Since this holds for all smooth fields ηa on M , we have (1.5.9).


1.6

38

Lie Derivatives

Let (M, C) be a fixed manifold, and let ξ a be a smooth vector field on M . The Lie derivative operator

£ξ associated with ξ a is a map from smooth tensor fields (on M ) to smooth tensor fields (on M ) of the ...ar ...ar same index structure. Roughly speaking, £ξ λba11...b represents the “rate of change” of the field λba11...b s s

relative to a standard of constancy determined by ξ a . We now have the tools in place to make this precise. (It is not important, but we write “£ξ ” rather than “£ξa ” to avoid the impression that the operator adds a new index. There is no chance for confusion since the object X in £X is always a contravariant vector field and the index it carries makes no difference.) ...ar Let λba11...b be a smooth field on M , and let p be a point in M . Further, let {Γt : U → Γt [U ]}t∈I be s

a local one–parameter group of diffeomorphisms generated by ξ a with p ∈ U . Here I is an open interval

of R, U is an open subset of M , and the maps Γt : U → Γt [U ] ⊆ M satisfy conditions (1) - (3) at the close of section 1.3. We set

...ar £ξ λba11...b s

|p

= lim

t→0

i 1h ...ar ...ar . (Γt )∗ (λba11...b ) |p − λba11...b |p s s t

...ar The right side limit is to be understood this way. We start with the tensor (λab11...b ) |Γt (p) at Γt (p), s ...ar carry it back to p with the pull back map (Γt )∗ , subtract (λba11...b ) , divide by t, and then take the limit s |p ...ar as t goes to 0. (That the limit exists, and that the resultant field £ξ λba11...b on M is smooth, follows s ...ar from proposition 1.3.3.) Note that we need to carry (λba11...b ) |Γt (p) back to p before comparing it with s a1 ...ar ...ar ...ar (λb1 ...bs )|p because the two tensors live in different spaces. The expression λba11...b is − λba11...b s |Γ(p) s |p

not well formed.

The following proposition lists several basic properties of Lie derivatives. (The proof is straightforward.) Proposition 1.6.1. The operator £ξ has the following properties. ab ab ab (1) It commutes with addition. E.g., £ξ (αab c + βc ) = £ξ (αc ) + £ξ (βc ).

(2) It satisfies the Leibniz rule with respect to exterior multiplication. ab ab For example, £ξ (αab c βdf ) = αc £ξ βdf + βdf £ξ αc .

(3) It commutes with the operation of index substitution. (4) It commutes with the operation of contraction. Problem 1.6.1. Show that £ξ δab = 0. (Hint: Recall that δab can be thought of as an index substitution operator, and make use of proposition 1.6.1.) Problem 1.6.2. Let η a be a smooth, non-vanishing field on M . Show that if £ξ (η a η b ) = 0, then £ξ η a = 0. Two cases are of special interest, namely Lie derivatives of scalar fields and of contravariant vector fields. We consider them in order.


39

Proposition 1.6.2. Let ξ a and α be smooth fields on M . Then £ξ (α) = ξ(α), i.e., at every point in M , £ξ α is just the ordinary directional derivative of α in the direction ξ a . Proof. Let p be any point in M , and let {Γt : U → Γt [U ]}t∈I be a local one–parameter group of diffeo-

morphisms generated by ξ a with p ∈ U . Since the curve γ : I → M defined by γ(t) = Γt (p) is an integral

curve of ξ a with initial point p, we have →

ξ|p (α) = γ |p (α) =

d d α ◦ Γt (p) |t=0 . (α ◦ γ)(0) = dt dt

But (Γt )∗ (α) = (α ◦ Γt ) for all t ∈ I. (Recall equation (1.5.6).) So we also have

i d 1 1h α ◦ Γt (p) − α ◦ Γ0 (p) = α ◦ Γt (p) |t=0 . (Γt )∗ (α) |p − α|p = lim £ξ α |p = lim t→0 t t→0 t dt

So (£ξ α)|p = ξ|p (α) at all points p in M .

We need a lemma for the second special case (Lie derivatives of contravariant vector fields). Lemma 1.6.3. Let ξ a be a smooth vector field on M , let p be a point in M and, once again, let {Γt : U → Γt [U ]}t∈I be a local one–parameter group of diffeomorphisms generated by ξ a with p ∈ U . Then, given any smooth scalar field α : M → R, there is a one–parameter family of smooth scalar fields

{ϕt }t∈I on U such that

(1) α ◦ Γt = α + t · ϕt for all t in I, and (2) ϕ0 = ξ(α). Proof. Consider the family of smooth scalar fields {ϕt }t∈I on U defined by setting Z 1 d ϕt (q) = α ◦ Γu (q) |u=t x dx du 0

for all t in I and q in U . We claim that it satisfies condition (1) and (2). First, for all t in I, Z 1 d t · ϕt (q) = α ◦ Γu (q) |u=t x t dx 0 du = =

Z

1 0

d α ◦ Γt x (q) dx dx

α ◦ Γt (q) − α ◦ Γ0 (q).

But Γ0 (q) = q for all q in U . So we have condition (1). Next, differentiation with respect to t yields t·

d d α ◦ Γt (q) . ϕt (q) + ϕt (q) = dt dt

Evaluating both sides at t = 0 gives us:

ϕ0 (q) =

d α ◦ Γt (q) |t=0 . dt


40

But now, since {Γt : U → Γt [U ]}t∈I is a local one–parameter group of diffeomorphisms generated by ξ a ,

the curve γ : I → O defined by γ(t) = Γt (q) is an integral curve of ξ a with initial value q. Thus, →

ξ|q (α) = γ |q (α) =

d d (α ◦ γ)| t=0 = α ◦ Γt (q) | t=0 . dt dt

So ϕ0 (q) = ξ|q (α) for arbitrary q in U . This is just condition (2).

Proposition 1.6.4. Let ξ a and λa be smooth vector fields on M . Then £ξ (λa ) = [ξ, λ]a , where [ξ, λ]a is the smooth (“commutator”) vector field on M whose action on a smooth scalar field α : M → R is

given by

[ξ, λ](α) = ξ(λ(α)) − λ(ξ(α)). (Another remark about notation. One must make some decision about how to handle abstract indices when dealing with commutator vector fields. Depending on context, we shall write, for example, either “[ξ, λ]a ” or “[ξ, λ]”— but never “[ξ a , λa ]” or “[ξ a , λa ]a ”. Nothing of importance turns on this decision.) Proof. Let p be any point in M , and let {Γt : U → Γt [U ]}t∈I be a local one–parameter group of diffeo-

morphisms generated by ξ a with p ∈ U . Given a smooth scalar field α : M → R, let {ϕt }t∈I be a

one–parameter family of smooth scalar fields on U satisfying conditions (1) and (2) in the lemma. For

all t such that both t and −t are in I, we have [(Γt )∗ (λa )]|p (α) = λa |Γt (p) (α ◦ Γ−t ) = λa |Γt (p) (α − t · ϕ−t ). The first equality follows directly from (1.5.1) and the fact that (Γt )∗ = (Γ−t )∗ . The second follows from condition (1) of the lemma (with t replaced by −t). So £ξ λa

(α) |p

= = =

lim

t→0

lim

t→0

lim

t→0

i 1h (Γt )∗ (λa ) |p (α) − λa |p (α) t

1 a λ |Γt (p) (α) − λa |p (α) − lim λa |Γt (p) (ϕ−t ) t→0 t 1 a (λ (α))|Γt (p) − (λa (α))|p − λa |p (ϕ0 ) t

d a λ (α) ◦ Γt (p) |t=0 − λa|p (ϕ0 ). dt Now the first term on the right side of the final line is equal to ξ|p λ(α) . (The argument is the same as =

used in the final stage of the proof of the lemma.) And ϕ0 = ξ(α), by condition (2) of the lemma. So £ξ λa

|p

(α) = ξ a |p (λ(α)) − λa |p (ξ(α)).

Since p and α are arbitrary, this establishes our claim.


41

Problem 1.6.3. Show that the set of smooth contravariant vector fields on M forms a “Lie algebra” under the bracket operation (defined in the preceding proposition), i.e., show that for all smooth vector fields ξ, η, λ on M , [ξ, η] = −[η, ξ]

and

λ, [ξ, η] + η, [λ, ξ] + ξ, [η, λ] = 0.

Problem 1.6.4. Show that for all smooth vector fields ξ a , η a on M , and all smooth scalar fields α on M,

£(α ξ) η a = α £ξ η a − £η α ξ a .

Problem 1.6.5. One might be tempted to take a smooth tensor field to be “constant” if its Lie derivatives with respect to all smooth vector fields are zero. But this idea does not work. Any contravariant vector field that was “constant” in this sense would have to vanish everywhere. Prove this. Problem 1.6.6. Show that for all smooth vector fields ξ a , η a , and all smooth tensor fields αa...b c...d , a...b £ξ £η − £η £ξ αa...b c...d = £θ αc...d

where θa is the field £ξ η a . It follows that £ξ and £η commute iff [ξ, η] = 0. (Hint: First prove the assertion, in order, for scalar fields α and contravariant fields αa . It will then be clear how to continue with covariant fields αa and arbitrary tensor fields αa...b c...d .) Although it is important to know how Lie derivatives are defined, in practice one rarely makes direct reference to the definition. Instead, one invokes propositions 1.6.1, 1.6.2, and 1.6.4. In fact, Lie derivatives can be fully characterized in terms of the properties listed there. Proposition 1.6.5. Let ξ a be a smooth vector field on M . Let D be an operator taking smooth tensor

fields on M to smooth tensor fields on M of the same index structure that satisfies the following three conditions. (1) For all smooth scalar fields α on M , D(α) = ξ(α). (2) For all smooth vector fields λa on M , D(λa ) = [ξ, λ]a . (3) D commutes with the operations of addition, index substitution, and contraction; it further satisfies the Leibniz rule with respect to tensor multiplication.

Then D = £ξ , i.e., D and £ξ have the same action on all smooth tensor fields. Proof. We are assuming outright that D and £ξ have the same action on scalar field and contravariant vector fields. We must show that (3) induces agreement on tensor fields of all other index structures. Consider, first, the case of a field γa . Given any smooth field λa on M , we must have D(γa λa ) =

ξ(γa λa ) = £ξ (γa λa ) by (1). Hence, by (3),

γa D(λa ) + D(γa )λa = γa £ξ (λa ) + £ξ (γa )λa .

42


But D(λa ) = £ξ (λa ) by (2). So, for arbitrary smooth fields λa on M , D(γa ) = £ξ (γa ).

D(γa ) − £ξ (γa ) λa = 0. Thus,

...ar We can now jump to the general case of a smooth tensor field λba11...b on M . We do so with an argument s

that is much like the one just used to handle the case of covariant vector fields. Let λb1 , ..., ρbs , µa1 , ..., νar ...ar b1 be arbitrary smooth fields on M , and consider the scalar field α = λba11...b λ ... ρbs µa1 ... νar . By (1), s

D(α) = £ξ (α). We can expand the terms D(α) and £ξ (α) using the fact that both operators, D and £ξ ,

satisfy the Leibniz rule. The result will be an equation with r + s + 1 terms on each side. The terms will

agree completely, except that where D appears on the left, £ξ will appear on the right. In r + s terms, the operator (D or £ξ ) will act on a vector field. So all these terms will cancel since D and £ξ agree in their action on contravariant and covariant vector fields. For example, the terms ...ar λba11...b D(λb1 ) ... ρbs µa1 ... νar s

...ar and λba11...b £ξ λb1 ... ρbs µa1 ... νar s

will cancel since D(λb1 ) = £ξ λb1 . So we may conclude that

b1 bs ...ar ...ar D λba11...b − £ξ λba11...b λ ...ρ µa1 ...νar = 0 s s

...ar . for all smooth fields λb1 , ..., ρbs , µa1 , ..., νar on M . Thus D and £ξ agree in their action on λba11...b s ...ar ...ar We record one more fact for future reference. For any smooth field λba11...b , £ξ λba11...b is supposed s s ...ar to represent the “rate of change” of the field λba11...b relative to a standard of constancy determined by s ...ar (the flow maps associated with) ξ a . So one would expect that £ξ λba11...b vanishes (everywhere) iff those s ...ar flow maps preserve λba11...b . We make the claim precise in the following proposition. The only slightly s

delicate matter is the need to keep track of the domains of definition of the local flow maps. ...ar Proposition 1.6.6. Let ξ a and λba11...b be smooth fields on M . Then the following conditions are equivs

alent. ...ar (1) £ξ λba11...b = 0 (everywhere on M ). s

(2) For all local one–parameter groups of diffeomorphisms {Γt : U → Γt [U ]}t∈I generated by ξ a , and all ...ar ...ar t ∈ I, (Γt )∗ (λba11...b ) = λba11...b . s s

Proof. The proof is essentially the same no matter what the index structure of the field under consideration. So, for convenience, we work with a field λab . One direction ((2) ⇒ (1)) is immediate. Let

{Γt : U → M }t∈I be any local one–parameter group of diffeomorphisms determined by ξ a , and let p be any point in U . If (2) holds, then, in particular, (Γt )∗ (λab ) |p = λab |p for all t ∈ I. Hence £λ λab

|p

= lim

t→0

i 1h 1 (Γt )∗ (λab ) |p − λab |p = lim [λab |p − λab |p ] = 0. t→0 t t

The converse requires just a bit more work. Suppose that (1) holds. Let {Γt : U → M }t∈I be any local one–parameter group of diffeomorphisms determined by ξ a , and let p be any point in U . Further, let ηa

43


and ρb and be any two vectors at p, and let f : I → R be the smooth map defined by f (t) =

Γt )∗ (λab ) ηa ρb . |p

We show that f ′ (t) = 0 for all t ∈ I. This will suffice. For then it will follow that f is constant, i.e., Γt )∗ (λab ) |p ηa ρb = Γ0 )∗ (λab ) |p ηa ρb = λab |p ηa ρb for all t ∈ I. Hence, since ηa and ρa are arbitrary vectors at p, it will follow that (Γt )∗ (λab ) |p = λab |p for all t ∈ I, as needed. So let t be any number in I. Then we have f ′ (t) =

lim

s→0

i 1h (Γt+s )∗ (λab ) |p ηa ρb − (Γt )∗ (λab ) |p ηa ρb . s

Now suppose s is sufficiently small in absolute value that {s, t + s} ⊆ I and Γs (p) ∈ U . Then Γt+s (p) =

(Γt ◦ Γs )(p). (Recall condition (2) in the final paragraph of section 1.3.) Hence, for all such s, we have

and

(Γt )∗ (λab ) |p ηa ρb = λab |Γt (p) (Γt )∗ (ηa ) |Γt (p) (Γt )∗ (ρb ) |Γt (p) (Γt+s )∗ (λab ) |p ηa ρb = (Γs )∗ (λab ) |Γt (p) (Γt )∗ (ηa ) |Γt (p) (Γt )∗ (ρb ) |Γt (p) .

So, substituting into our expression for f ′ (t), we have 1 ∗ a a ′ (Γs ) (λb ) |Γt (p) − λb |Γt (p) (Γt )∗ (ηa ) |Γt (p) (Γt )∗ (ρb ) |Γt (p) f (t) = lim s→0 s = £ξ λab |Γt (p) (Γt )∗ (ηa ) |Γt (p) (Γt )∗ (ρb ) |Γt (p) . Since £ξ λab = 0 everywhere, we may conclude that f ′ (t) = 0.


1.7

44

Derivative Operators and Geodesics

We have already introduced one kind of derivative operator, namely £λ , associated with a smooth contravariant vector field λa . In this section, we discuss a different kind. It is, in a sense, a generalization of the gradient operator ∇ that one encounters in standard vector analysis on Rn . ...ar Let M be a manifold, and let ∇ be a map that acts on pairs (c, αba11...b ), where the second is a smooth s

tensor field on M and the first is an abstract index distinct from a1 , ..., ar , b1 , ..., bs , and associates with

...ar them a smooth tensor field ∇c αba11...b on M in which c appears as a covariant index. (Given any one s

...ar ...ar index c, we understand ∇c to be the operator that takes the field αba11...b to the field ∇c αba11...b .) We s s

say that ∇ is a (covariant) derivative operator on M if it satisfies the following conditions. (DO1) ∇ commutes with addition on tensor fields. ba ba = ∇n αab For example, ∇n αab c + ∇n βc . c + βc

(DO2) ∇ satisfies the Leibniz rule with respect to tensor multiplication. ab ab For example, ∇n αab c ξf d = αc ∇n ξf d + ∇n αc ξf d . (DO3) ∇ commutes with index substitution.

For example, the result of applying (a → d) index substitution to αab c and applying ∇n is the same

as that arising from applying (a → d) substitution to ∇n αab c . Furthermore, the result of applying (n → m) index substitution to ∇n αab is the same as that arising from applying ∇m to αab c c .

(DO4) ∇ commutes with contraction.

For example, the result of applying (a, c) contraction to ∇n αab is the same as that arising from c applying ∇n to αab a .

(DO5) For all smooth scalar fields α and all smooth vector fields ξ n , ξ n ∇n α = ξ(α). (DO6) For all (distinct) indices a and b, ∇a ∇b α = ∇b ∇a α. The first four conditions should seem relatively innocuous. DO5 is suggested by the situation in ordinary vector analysis on Rn . There the directional derivative of α in the direction ξ is given by ξ · ∇α. (Recall

equation (1.2.1).) We want to interpret ξ n ∇n α as the analog of ξ · ∇ α. So we set ξ n ∇n α equal to

the (generalized) directional derivative ξ(α). DO6 is a bit more delicate. One can imagine strengthening the condition to require that ∇a and ∇b commute on all tensor fields. This leads to the class of “flat”

derivative operators, and is far too restrictive for our purposes. One can also imagine dropping the

condition altogether. This leads to the larger class of “derivative operators with torsion”. It will be clear later why we have included DO6. (The derivative operators determined by metrics are necessarily torsion free.)

45


Some authors refer to the associated maps ∇a as “derivative operators”, rather than reserving that

term for the map ∇ itself. We shall do so as well on occasion.

Having defined derivative operators, we can now pose the question of their existence and uniqueness on manifolds. Concerning existence, one has the following basic result (Geroch [23, appendix]). Proposition 1.7.1. A connected manifold admits a derivative operator iff a countable subset of the manifold’s charts suffice to cover it. The restriction to connected manifolds here is harmless since, clearly, a manifold admits a derivative operator iff each of its components does. Practically all the manifolds one ever deals with in differential geometry satisfy the stated countable cover condition. Indeed, one has to work hard to find a manifold that does not. So proposition 1.7.1 has the force of a strong existence theorem. (And, of course, it implies that all manifolds admit derivative operators locally.) The question of uniqueness is easier to deal with, and we give a complete answer. But first a lemma is needed. Lemma 1.7.2. Let ∇ be a derivative operator on the n-dimensional manifold M , and let ξb be a co-vector at the point p. Then there is a smooth scalar field α in S(p) such that ξb = (∇b α)|p .

Proof. Here we use coordinates as in section 1.2. Suppose (U, ϕ) is a chart on M with p ∈ U , and u1 , ..., un →

→

are the corresponding coordinate maps on U . The coordinate curve tangent vectors γ 1|p , ..., γ n|p form a 1

2

→

n

j

basis for Mp . Let {β, β, ..., β} be a dual basis. Then ( γ i |p ) β = δij for all i and j in {1, ..., n}, and there n i P i 1 n c β b . Now we define a smooth scalar field α : U → R by exist real numbers c, ..., c such that ξb = setting α(q) =

n P

i=1

i=1

i

i

c u (q). We claim ξb = (∇b α)|p .

We must show that η b ξb = η b (∇b α)|p holds for arbitrary vectors η b at p. Let η b = one such. Then, by DO5, and the fact that

→ γ i |p (uj )

η b (∇b α)|p = η(α)|p =

= δij (recall (1.2.6)), we have  ! n n n X X Xj i i → i d c. d ( γ i |p )  c uj  = i=1

Since we also have η b ξb =

i=1

i

→

d γ i |p

b

be

i=1

j=1

 ! n n n X X Xj j b i → i i  d γ i |p d c, c β b = i=1

we are done.

n P

j=1

i=1

Proposition 1.7.3. Let ∇ and ∇′ be derivative operators on the manifold M . Then there exists a

a smooth symmetric tensor field Cbc on M that satisfies the following condition for all smooth tensor

46


...ar fields αba11...b on M : s

...ar = ∇′m − ∇m αba11...b s

a ...a

n

...ar a1 ...ar 1 r−1 ar n a1 n 2 ...ar . Cmn + ... + αba11...b αnb − αna Cmb Cmb b1 ...bs Cmn − ... − αb1 ...bs 1 s s−1 n 2 ...bs

(1.7.1)

m Conversely, given any derivative operator ∇ on M and any smooth symmetric tensor field Cab on M , if

∇′ is defined by (1.7.1), then ∇′ is also a derivative operator on M .

...ar (To get a grip on (1.7.1), note that for each index in αbas1...b there is a corresponding term on the right. s

That term carries a + or − depending on whether the index is a subscript or superscript. In that term,

a the index is contracted into Cbc .)

Proof. Let ∇ and ∇′ be derivative operators on M . Note first that given any smooth scalar field α on

M , ∇′a α = ∇a α. (This follows from the fact that given any vector ξ a at any point in M , ξ a ∇′a α = ξ(α) = ξ a ∇a α.)

Next we claim that given any smooth co-vector field γb on M , if γb = 0 at a point p, then ∇′a γb = ∇a γb

at p. To see this, let ξ b be any smooth field on M and consider the scalar field γb ξ b . We have 0 = ∇′a − ∇a γb ξ b = γb ∇′a − ∇a ξ b + ξ b ∇′a − ∇a γb everywhere. So, in particular, we have 0 = ξ b ∇′a − ∇a γb at p. Since this is true for arbitrary ξ a , it must be the case that ∇′a − ∇a γb = 0

as claimed.

It follows from the claim that given any smooth field αb on M , the value of 1

∇′m − ∇m αb at a point 2

p is determined solely by the value of αb itself at p. For suppose that αb and αb agree at p. Then the 2 1 1 2 claim is applicable to αb − αb , and therefore ∇′m − ∇m αb = ∇′m − ∇m αb at p.) 0

a Now we define a tensor field Cbc . Given any point p and a vector αa at p, we set

n 0 Cmb αn = ∇′m − ∇m αb 0

where αb is any smooth field on M that assumes the value αb at p. (Our preliminary work shows that n n satisfies Cmb αn = ∇′m − ∇m αb the choice of αb makes no difference.) It follows immediately that Cmb

for all smooth fields αb and, therefore, is smooth itself.

a Cbc is symmetric. To see this, consider any smooth scalar field α on M . Since ∇′n α = ∇n α, it n follows that Cmb ∇n α = ∇′m − ∇m ∇b α = ∇′m ∇b α − ∇m ∇b α = ∇′m ∇′b α − ∇m ∇b α. So, by DO6, we

n n may conclude that Cmb ∇n α = Cbm ∇n α. But by our lemma, all covariant vectors at a point can be

a a . = Ccb realized in the form ∇n α for some scalar field α. So we have Cbc

a Next we show that Cbc satisfies condition (1.7.1). This involves a now familiar sequential form of

argument — from scalar fields, to vector fields, to arbitrary tensor fields. We have already seen that all a derivative operators agree on scalar fields. And it follows directly from our definition of Cbc that (1.7.1)

47


holds for covariant vector fields. So let ξ a be an arbitrary smooth contravariant field on M . Then, given any smooth field γa on M , 0

∇′a − ∇a ξ b γb = ξ b ∇′a − ∇a γb + γb ∇′a − ∇a ξ b d = ξ b Cab γd + γb ∇′a − ∇a ξ b h i d = ξ b Cab + ∇′a − ∇a ξ d γd .

=

d , as required by (1.7.1). ∇′a − ∇a ξ d = −ξ b Cab To check (1.7.1) for tensor fields αabc , one expands 0 = ∇′m − ∇m αabc ξ b λc ηa for arbitrary fields ξ b , λc , ηa and uses the known expressions for ∇′m − ∇m ξ b , ∇′m − ∇m λc , and ∇′m − ∇m ηa . The Since this holds for all smooth fields γa , it follows that

calculation is straightforward. Tensor fields of arbitrary index structure can be handled similarly. The second half of the proposition is also straightforward.

a It is worth noting that condition DO6 entered the proof only in the demonstration that Cbc must be a symmetric. If in the statement of the proposition one drops the requirement of symmetry on Cbc , then

one has the appropriate formulation for derivative operators with torsion. a In what follows, if ∇′ and ∇ are derivative operators on a manifold that, together with the field Cbc , ′ a ′ a satisfy condition (1.7.1), then we shall write ∇ = ∇, Cbc . Clearly this is equivalent to ∇ = ∇ , −Cbc .

We have introduced two kinds of derivative operators. The next proposition shows how the action of

one can be expressed in terms of the other. Proposition 1.7.4. Suppose ∇ is a derivative operator on the manifold M , and λa is a smooth vector ...ar field on M . Then for all smooth fields αba11...b on M , we have: s ...ar £λ αba11...b s

=

...ar ...ar a1 ...ar λn ∇n αba11...b + αnb ∇b1 λn + ... + αba11...b ∇bs λn s s−1 n 2 ...bs 2 ...ar −αbna ∇n λa1 − ... − 1 ...bs

(1.7.2)

a ...a n αb11...bsr−1 ∇n λar .

...ar ...ar (Condition (1.7.2), of course, resembles (1.7.1) above. The difference £λ αba11...b − λn ∇n αba11...b is a s s ...ar sum of terms, one for each index in αba11...b . s

The terms carry a + or − depending on whether the

associated index is a subscript or a superscript. Each term is contracted with ∇a λb .)

Proof. The proof is another simple sequential argument, like the one used in the preceding proof. (Note that we shall not need to invoke the definition of Lie derivatives. It will suffice to make use of the properties collected in propositions 1.6.1 and 1.6.4.) First of all, trivially, if α is a smooth scalar field, then £λ α = λ(α) = λn ∇n α. Next, suppose ξ a is

a smooth vector field. Then for arbitrary smooth scalar fields α, we have, by proposition 1.6.4, £λ ξ (α)

= = =

λ ξ(α) − ξ λ(α) = λ ξ a ∇a α − ξ λa ∇a α λb ∇b ξ a ∇a α − ξ b ∇b λa ∇a α λb ξ a ∇b ∇a α + λb ∇b ξ a ∇a α − ξ b λa ∇b ∇a α − ξ b ∇b λa ∇a α.


48

The first and third term of the last line cancel each other by DO6. So we have £λ ξ (α) = λb ∇b ξ a − ξ b ∇b λa ∇a α = λb ∇b ξ a − ξ b ∇b λa (α).

Since α is arbitrary, it follows that

£λ ξ a = λb ∇b ξ a − ξ b ∇b λa , as required by equation (1.7.2). Next let αa be a smooth covariant vector field. Then for arbitrary smooth fields ξ a , £λ (αa ξ a ) = =

αa £λ ξ a + ξ a £λ αa αa λb ∇b ξ a − ξ b ∇b λa + ξ a £λ αa .

Here we have used both the fact that £λ satisfies the Leibniz condition and our previous expression for £λ ξ a . But we also have

Therefore,

£λ αa ξ a = λb ∇b αa ξ a = λb αa ∇b ξ a + λb ξ a ∇b αa .

Since ξ a is arbitrary, we have

ξ a £λ αa = ξ a λb ∇b αa + αb ∇a λb . £λ αa = λb ∇b αa + αb ∇a λb ,

as required by equation (1.7.2). Continuing this way, we can verify (1.7.2) for tensor fields of arbitrary index structure. Problem 1.7.1. Show that if ∇ is a derivative operator on a manifold, then ∇n δab = 0. With the notion of a derivative operator in hand, we can now introduce the idea of “parallel transport” of tensors along curves. Suppose M is a manifold with derivative operator ∇. The directional derivative of a scalar field α at

p in the direction ξ a , we know, is given by ξ n ∇n α. Generalizing now, we take the directional derivative ...am of a smooth field αba11...b at p in the direction ξ a (with respect to ∇) to be r ...am ξ n ∇n αba11...b . r

...am Furthermore, if γ : I → M is a smooth curve with tangent field ξ a , we say that αba11...b is constant r

...am along γ (with respect to ∇) if ξ n ∇n αba11...b = 0. r

Derivative operators are sometimes called “connections” (or “affine connections”). That is because, in a sense, they “connect” the tangent spaces of points “infinitesimally close” to one another, i.e., they provide a standard of identity for vectors at distinct, but “infinitesimally close”, points.

49


So far, our tensor fields have always been defined over an entire manifold or — this amounts to the same thing — to open subsets of a manifold. It is useful also to consider tensor fields defined on curves. Suppose γ : I → M is a smooth curve on the manifold M . A tensor field (of a given index structure) on

γ is just a map that assigns to each s in I a tensor of that index structure at γ(s). (Note that this is not

quite the same as assigning a tensor of that index structure to each point in γ[I], since we are not here excluding the possibility that the curve may cross itself, i.e., that γ(s1 ) = γ(s2 ) for distinct s1 and s2 in I. We do not want to insist in such a case that the tensor assigned to s1 is the same as the one assigned to s2 .) So, for example, the tangent field to γ counts as a tensor field on γ. It is clear what the appropriate criterion of smoothness is for tensor fields on γ. A scalar field on γ is just a map α : I → R. So we certainly understand what it means for it to be smooth. We take a vector

field ξ a on γ to be smooth if, for all smooth scalar fields α on M , ξ(α) is a smooth scalar field on γ.

Next, we take a co-vector field µa on γ to be smooth if, for all smooth fields ξ a on M , ξ a µa is a smooth scalar field on γ. One can continue in this way following the usual pattern. Note that the tangent vector field to any smooth curve qualifies as smooth. ...am Now suppose that γ : I → M is a smooth curve on the manifold M with tangent field ξ a , αba11...b is r

...am a smooth field on γ, and ∇ is a derivative operator on M . We cannot meaningfully apply ∇ to αba11...b . r

...am But we can make sense of the directional derivative field ξ n ∇n αba11...b on γ. We can do so using the r

following proposition.

Proposition 1.7.5. Suppose ∇ is a derivative operator on the manifold M and γ : I → M is a smooth curve with tangent field ξ a . Then there is a unique operator

...am ...am αba11...b 7→ D αba11...b r r

taking smooth tensor fields on γ to smooth tensor fields on γ of the same index structure that satisfies the following conditions. (1) D commutes with the operations of addition, index substitution, and contraction; it further satisfies the Leibniz rule with respect to tensor multiplication.

dα . ds ...am (3) Let s 7→ αba11...b (s) be a smooth tensor field on γ. Suppose there is an open set O and a smooth r (2) For all smooth scalar fields s 7→ α(s) on γ, D(α) =

...am field α eba11...b on O such that, for all s in some open interval I ′ ⊆ I, r ...am ...am Then, for all s in I ′ , D αba11...b (s) = ξ n ∇n α eba11...b |γ(s) . r r

...am ...am αba11...b (s) = α eba11...b . r r |γ(s)

...am Proof. Suppose first that D satisfies the stated conditions, and suppose αba11...b r

field on γ. We shall derive an explicit expression for

...am D(αba11...b ) r

is a smooth tensor

in terms of a local coordinate chart.

This will show that there can be, at most, one D satisfying the stated conditions. To avoid drowning in

indices, we shall work with a representative case – a smooth field αab – but it will be clear how to adapt

the argument to fields with other index structures.

50


Suppose our background manifold (M, C) has dimension n. Let s be any point in I, and let (U, ϕ) be i

an n-chart in C whose domain U contains γ(s). For all i ∈ {1, ..., n}, let η a be the smooth coordinate→

curve tangent field ( γ i )a on U . We know that the fields 1

2

1

2

n

η a , η a , ..., η a

form a basis for the tangent

n

space at every point in U . Let µa , µa , ..., µa be corresponding smooth co-vector fields on U that form a dual basis at every point. Now let αab be a smooth field on γ. We can certainly express it in terms of these basis and co-basis fields. That is, we can find an open subinterval I ′ ⊆ I containing s, and smooth ij

functions α : I ′ → R such that, at all points s in I ′ , αab (s) =

n X

ij

i

j

α(s) (η a )|γ(s) (µb )|γ(s) .

i,j = 1 ij

i

j

i

j

Here, of course, α = αab η b µa . We can construe the restrictions of η a and µb to γ[I ′ ] as smooth fields on (a restricted segment of) γ. It follows, therefore, that at all points in I ′ , D(αab ) =

n X

i,j = 1

n h i ij i j X ij ij i j i j D(α) (η a µb ) + α D(η a µb ) . D α η a µb = i,j = 1

Here we have just used the fact that D commutes with tensor addition and satisfies the Leibniz rule

(and suppressed explicit reference to the evaluation point s). But now it follows from conditions (2) and ij

ij

dα i j i j , and D(η a µb ) = ξ n ∇n (η a µb ). So, we have our promised explicit ds   ij n X d α ij n ia j ia j a  D(αb ) = (η µb ) + α ξ ∇n (η µb ) . ds i,j = 1

(3), respectively, that D(α) =

expression for D(αab ):

To show existence, finally, it suffices to check that the operator D defined by this expression (and

the counterpart expressions for fields with other index structures) satisfies all three conditions in the proposition.

...am Under the stated conditions of the proposition, we can now understand ξ n ∇n αba11...b to be the r ...am smooth field on γ given by D αba11...b . Note that condition (3) in the proposition just makes precise r

...am the requirement that ξ n ∇n αba11...b r

...am is “what it should be” in the case where αba11...b r

arises as the

restriction to γ[I] of some smooth tensor field defined on an open set. We have already said what it means for a tensor field defined on an open set to be “constant” along ...am a curve γ with tangent field ξ a . We can now extend that notion to fields αba11...b r

itself. The defining condition,

...am ξ n ∇n αba11...b r

defined only on γ

= 0 carries over intact.

The fundamental fact about constant fields on curves is the following. Proposition 1.7.6. Given a manifold M , a derivative operator ∇ on M , a smooth curve γ : I → M ,

...am ...am and a tensor α ba11...b at some point γ(s), there is a unique smooth tensor field αba11...b on γ that is r r 0

...am constant (with respect to ∇) and assumes the value α ba11...b at s. r 0

51


...am When the conditions of the proposition are realized we say that αba11...b results from parallel transport r 0

a1 ...am

of αb1 ...br

along γ (with respect to ∇).

Finally, we introduce “geodesics”. We say that a smooth curve γ : I → M is a geodesic (with respect

to ∇) if its tangent vector field ξ a is constant along γ, i.e., if ξ b ∇b ξ a = 0. The basic existence and

uniqueness theorem for geodesics is the following. (In what follows, we shall drop the qualification “with respect to ∇” except in contexts where doing so might lead to ambiguity.)

Proposition 1.7.7. Given a manifold M , a derivative operator ∇ on M , a point p in M , and a vector →

ξ a at p, there is a unique geodesic γ : I → M with γ(0) = p and ξ = γ |p that satisfies the following →

maximality condition: if γ ′ : I ′ → M is also a geodesic with γ ′ (0) = p and γ ′ |p = ξ, then I ′ ⊆ I and γ ′ (s) = γ(s) for all s ∈ I ′ .

To prove propositions 1.7.6 and 1.7.7, one formulates the assertions in terms of local coordinates and then invokes the fundamental existence and uniqueness theorem for ordinary differential equations. A derivative operator determines a class of geodesics. It turns out that a derivative operator is actually fully characterized by its associated geodesics. This will be important later in our discussion of relativity theory. Proposition 1.7.8. Suppose ∇ and ∇′ are both derivative operators on the manifold M . Further suppose

that ∇ and ∇′ admit the same geodesics (i.e., for all smooth curves γ : I → M , γ is a geodesic with

respect to ∇ iff it is a geodesic with respect to ∇′ ). Then ∇′ = ∇.

Proof. The argument provides a good example of how proposition 1.7.3 is used. Given ∇ and ∇′ , there

a a a ). It will suffice to show that Cbc on M such that ∇′ = (∇, Cbc must exist a smooth symmetric field Cbc

vanishes everywhere.

0

Given an arbitrary point p and an arbitrary vector ξ a at p, there is a geodesic γ with respect to ∇ 0

that passes through p and has tangent ξ a at p. Let ξ a be the tangent vector field of γ. Then we have ξ n ∇n ξ a = 0. By our hypothesis, γ must also be a geodesic with respect to ∇′ . So ξ n ∇′n ξ a = 0 too.

a b Now, since ∇′a = (∇a , Cbc ), we have ∇′ a λb = ∇a λb − Can λn for all smooth fields λa . So, in particular,

we have

b 0 = ξ a ∇′a ξ b = ξ a ∇a ξ b −Can ξa ξn | {z } =0

0

0

0

b b ξn ξa = 0 at p. But ξa and p were arbitrary, and Can is symmetric. at all points on the image of γ. So Can a So, by proposition 1.4.3, Cbc must vanish everywhere.

The property of being a geodesic is not preserved under reparametrization of curves. The situation is as follows.

52


Proposition 1.7.9. Suppose M is a manifold with derivative operator ∇, and γ : I → M is a smooth curve with tangent field ξ a . Then γ can be reparametrized so as to be a geodesic (i.e., there is a diffeo-

morphism α : I ′ → I of some interval I ′ onto I such that γ ′ = γ ◦ α is a geodesic) iff ξ n ∇n ξ a = f ξ a

for some smooth scalar field f on γ. Furthermore, if γ is a non-trivial geodesic (i.e., a geodesic with non-vanishing tangent field), then the reparametrized curve γ ′ = γ ◦ α is a geodesic iff α is linear. Proof. Suppose α : I ′ → I is a diffeomorphism and ξ ′ is the tangent field to γ ′ = γ ◦ α : I ′ → M . dα dα Set t = α(s). By the chain rule, we have ξ ′ = ξ . (This abbreviates ξ ′ |γ ′ (s) = ξ|γ(α(s)) (s). Recall ds ds dα as a smooth scalar field on γ – it (1.3.1) in the proof of proposition 1.3.2.) Now we can construe ds dα dα assigns to s the number (s) at the point γ(s) – and we can make sense of the rate of change ξ n ∇n . ds ds So we have 2 dα dα a dα dα dα n n a ′n ′a ξ a ξ n ∇n ∇n ( ξ ∇n ξ + ξ ξ )= . ξ ∇n ξ = ds ds ds ds ds Now

dα (s) 6= 0 for all s in I ′ , since α is a diffeomorphism. So, by the chain rule again, ds −1 dα d2 α dα d dα n ξ ∇n = 2 . = ds dt ds ds ds

It follows that ξ

′n

∇n ξ

′a

=

dα ds

2

ξ n ∇n ξ a +

d2 α a ξ . ds2

(1.7.3)

Both our claims follow from this last equation. First, γ ′ is a geodesic, i.e., ξ ′n ∇n ξ ′a = 0 iff ξ n ∇n ξ a = −2 d2 α dα . Second, if γ is a geodesic (i.e., if ξ n ∇n ξ a = 0), then γ ′ is also f ξ a , where f = − 2 ds ds d2 α a a geodesic iff ξ = 0. On the assumption that ξ a is non-vanishing, the latter conditions holds iff ds2 d2 α = 0, i.e., α is linear. ds2 We know that a derivative operator is determined by its associated class of geodesics. Let us now consider a different question. Suppose one does not know which (parametrized) curves are geodesics, but only which ordered point sets on a manifold are the images of geodesics. To what extent does that partial information allow one to determine the derivative operator? We answer the question in the next proposition. Let us say that two derivative operators ∇ and ∇′ on a manifold are projectively equivalent

if they admit the same geodesics up to reparametrization (i.e., if any curve can be reparametrized as to be a geodesic with respect to ∇ iff it can be reparametrized so as to be a geodesic with respect to ∇′ ).

a Proposition 1.7.10. Suppose ∇ and ∇′ are derivative operators on a manifold M and ∇′ = ∇, Cbc .

Then ∇ and ∇′ are projectively equivalent iff there is a smooth field ϕc on M such that a Cbc = δba ϕc + δca ϕb .

53


Proof. Suppose first that there does exist such a field ϕc . Further suppose that γ is an arbitrary smooth curve with tangent field ξ a . Then ξ n ∇′n ξ a

a a ϕn ξ n ξ m ξ m = ξ n ∇n ξ a − δna ϕm + δm = ξ n ∇n ξ a − Cnm = ξ n ∇n ξ a − 2 ξ a (ϕm ξ m ).

It follows by the first part of proposition 1.7.9 that γ can be reparametrized so as to be a geodesic with respect to ∇ iff it can be reparametrized so as to be a geodesic with respect to ∇′ . Conversely, suppose that ∇ and ∇′ are projectively equivalent. We show there is a smooth field ϕc on

a M such that Cbc = δba ϕc + δca ϕb . Let γ be an arbitrary geodesic with respect to ∇ with tangent field

ξ a . Then ξ n ∇n ξ a = 0 and ξ n ∇′n ξ a = f ξ a for some smooth field f on γ. (Here again we use the first part of proposition 1.7.9.) It follows that

Therefore,

a b c a c ξ ξ . ξ = −Cbc f ξ a = ξ b ∇b ξ a − Cbc

d a b c a d ξ ξ ξ = 0. This can be expressed as ξ − Cbc Cbc

Now let ϕad bcr be the field

a d d a b c r Cbc δr − Cbc δr ξ ξ ξ = 0.

a d d a Cbc δr − Cbc δr . Symmetrizing on the indices b, c, r, we have ad ξ b ξ c ξ r = 0. ϕ(bcr)

ad Since this equation must hold for all choices of γ, and hence all vectors ξ (at all points), and since ϕ(bcr) ad is symmetric in b, c, r, it follows from proposition 1.4.3 that ϕ(bcr) = 0. Therefore, using the fact that a is itself symmetric, Cbc a d d a a d d a a d d a Cbc δr − Cbc δr + Crb δc − Crb δc + Ccr δb − Ccr δb = 0.

Now suppose n is the dimension of our underlying manifold. Then (r, d) contraction yields a a a d a a d a n Cbc − Cbc + Ccb − Cdb δc + Ccb − Ccd δb = 0. a d d . If we set ϕc = Thus, (n + 1) Cbc = δba Ccd + δca Cbd

1 C d , this can be expressed as n + 1 cd

a Cbc = δba ϕc + δca ϕb .

We close this section with a few remarks about the “exterior derivative operator” and about “coordinate derivative operators” (associated with particular charts on a manifold). An m-form (for m ≥ 0) on a manifold M is a tensor field on M with m covariant indices that is

anti-symmetric, i.e., a tensor field of the form αa1 ...am where αa1 ...am = α[a1 ...am ] . (Scalar fields qualify as 0-forms.)

54


Suppose αa1 ...am is a smooth m-form on M , and c is an index distinct from a1 , ..., am . Then, given any covariant derivative operator ∇, ∇[c αa1 ...am ] qualifies as a smooth (m + 1)-form on M . It turns out

that this field is independent of the choice of derivative operator ∇. (See problem 1.7.2.) In this way, we arrive at an operator d (the exterior derivative operator) that acts on pairs (c, αa1 ...am ) and satisfies dc αa1 ...am = ∇[c αa1 ...am ]

(1.7.4)

for all choices of ∇. So, in particular, we have da α = ∇a α for all smooth scalar fields α. We have

db αa = ∇[b αa] = 12 (∇b αa − ∇a αb ) for all smooth co-vector fields αa , and so forth.

One can certainly introduce the exterior derivative operator directly, without reference to covariant derivative operators. Most books do so. But there is no loss in proceeding as we have since covariant derivative operators always exist locally on manifolds, and local existence is all that is needed for our characterization. Officially, we are taking the exterior derivative operator d to be a map that acts on a pair of objects — an index and a smooth m-form (for some m or other). One might also use the term to refer to the associated map dx that assigns dx αa1 ...am to αa1 ...am . Some authors do so, and we shall too on occasion. Problem 1.7.2. Let ∇ and ∇′ be derivative operators on a manifold, and let αa1 ...an be a smooth n-form on it. Show that

∇[b αa1 ...an] = ∇′[b αa1 ...an ] . (Hint: Make use of proposition 1.7.3.) It is worth asking why we do not allow the exterior derivative operator to act on arbitrary smooth covariant tensor fields. The problem is not a failure to be well defined. (Note that given any smooth covariant field αa1 ...am , and any two derivative operators ∇ and ∇′ , it follows from problem 1.7.2 that ∇[b αa1 ...an ] = ∇[b α[a1 ...an ]] = ∇′[b α[a1 ...an ]] = ∇′[b αa1 ...an ] .) Rather, the problem is that we cannot both extend the application of the exterior derivative operator and have it satisfy the Leibniz rule — and presumably the latter is a requirement for any derivative-like operator. Here is the argument. Let αab be any smooth symmetric field. Then (if we allow ourselves to apply dn ), dn αab = ∇[n αab] = 0. Now let f be any smooth scalar field. By the same argument, we

have dn (f αab ) = 0. So, if the Leibniz rule obtains, we have

0 = dn (f αab ) = f (dn αab ) + (dn f ) αab = (dn f ) αab . But this is impossible since, given any point p, we always choose f and αab so that neither αab nor (dn f ) vanishes at p. We have introduced three types of derivative operator on manifolds. It is helpful to contrast them with respect to two features: the background geometric structure they presuppose (if any) and the types of

55


tensor fields to which they can be applied. One finds a trade-off of sorts. The exterior derivative operator da presupposes no background structure (beyond basic manifold structure). But it is only applicable to smooth m-forms (for some m or other). In contrast, the Lie derivative operator £ξ and the covariant derivative operator ∇a can both be applied to arbitrary smooth tensor fields. But the first presupposes

(i.e., is defined relative to) a smooth contravariant vector field ξ; and the latter can itself be thought of as a layer of geometric structure beyond pure manifold structure. (Another difference, of course, is that £ξ leaves intact the index structure of the tensor field on which it acts, whereas da and ∇a both add a as a covariant index.)

Let us now consider “coordinate differentials”. Let (U, ϕ) be an n-chart on the n-manifold (M, C),

and let ui : U → R (i = 1, ..., n) be the coordinate maps on U determined by ϕ. We know that the →

→

associated smooth coordinate-curve tangent fields γ 1 , ..., γ n form a basis for the tangent space at every

point in U . (Recall the discussion in section 1.2.) The notation ∂ ∂ , ... , ∂u1 ∂un is often used for these fields. And give any smooth scalar field f on U , the action of ∂f written, simply, as . Using this notation, we have, by (1.2.5), ∂ui ! → ∂ f ◦ ϕ−1 ∂f ∂ = (f ) = γ i |p (f ) = ∂ui |p ∂ui |p ∂xi ϕ(p)

for all p in U . In particular, if we take f to be uj , it follows from (1.2.6) that j → ∂u ∂ = (uj ) = γ i (uj ) = δij i i ∂u ∂u

∂ ∂ui

on f is often

(1.7.5)

(1.7.6)

at all points in U . Furthermore, if ∇ is a derivative operator on M , we have, by condition (DO5),

∂ ∂ui

a

(da f ) =

∂ ∂ui

a

(∇a f ) =

∂ ∂ui

(f ) =

∂f ∂ui

.

(1.7.7)

So, taking f to be uj once again, we have

∂ ∂ui

a

(da uj ) =

∂uj ∂ui

= δij .

This shows that the co-vectors (da u1 ), ..., (da un ) form a dual basis to

∂ ∂u1

a

, ... ,

∂ ∂un

a

at every point in U .

(1.7.8)

56


Many useful facts follow from the preceding lines. For example, it follows that the index substitution field δba can be expressed as δba =

∂ ∂u1

a

(db u1 ) + ... +

∂ ∂un

a

(db un ).

(1.7.9)

And it follows that, for all smooth scalar fields f on U , db f =

n X ∂f (db uj ). j ∂u j=1

(1.7.10)

(In both cases, the left- and right-side fields must be equal since they have the same action on the basis b b ∂ ∂ fields . Consider (1.7.10). We know from (1.7.7) that contraction with on the left side ∂ui ∂ui b ∂ ∂f ∂f ; and we know from (1.7.8) that contraction with on the right side yields yields ∂ui ∂ui ∂ui

as well.) If we were not using the abstract index notation, we would express (1.7.10) in the form n X ∂f dui . df = j ∂u j=1 Next we consider “coordinate derivative operators”. The basic fact is this.

Proposition 1.7.11. Let M be an n-manifold, let (U, ϕ) be an n-chart with non-empty domain on M (in the atlas that defines the manifold), and let ui : U → R (i = 1, ..., n) be the coordinate maps determined b ∂ by ϕ. Then there is a unique derivative operator ∇ on U such that ∇a = 0 for all i.2 ∂ui Proof. Uniqueness follows easily from proposition 1.7.3. Suppose ∇ and ∇′ are derivative operators on

a U with ∇′ = (∇, Cbc ). Then, for all i,

∇′a

∂ ∂ui

b

= ∇a

∂ ∂ui

b

b − Can

∂ ∂ui

n

.

b This, in turn, implies that Can

n ∂ = 0, for all i. ∂ui = 0. (Somewhat more generally, if two derivative operators agree in their

b So if ∇ and ∇′ both satisfy the stated condition, it must be the case that Can

action on a set of vector fields that span the tangent space at each point, the derivative operators must be equal.) We now establish existence by explicitly exhibiting a derivative operator ∇ on U that satisfies the

stated condition. First, given any smooth scalar field f on U , we take ∇a f to be the field on the right

side of (1.7.10). (We have no choice here, since da f = ∇a f for all derivative operators.) Next consider

any smooth tensor field on U that carries at least one abstract index. It can be expressed uniquely as 2 Here,

as usual, we have suppressed explicit reference to manifold atlases. We mean, of course, that (M, C) is an

n-manifold, (U, ϕ) is an n-chart in C, and ∇ is a derivative operator on the restricted manifold (U, C|U ).

57


a sum over the basis fields uniquely in the form

∂ ∂ui

a

and (da uj ). Consider an example. The field γcab can be expressed n X n X n X

γcab =

ijk

γ

i=1 j=1 k=1

We take the action of ∇m on γcab to be given by ∇m γcab

=

n X n X n X n X i=1 j=1 k=1 l=1

!

ijk

∂γ ∂ul

a

∂ ∂ui

∂ ∂ui

∂ ∂uj

a

b

(dc uk ).

∂ ∂uj

b

(dc uk )(dm ul ). ijk

Here we introduce a new summation variable l, take the partial derivative of the scalar field γ with respect to ul , and add (dm ul ) to the list of fields on the right. This prescription can be generalized. In every case, we the action of ∇m on a tensor field by first expressing the field as a sum over the determine a ∂ basis fields and (db uj ), and then generating a new sum (with m as a new covariant index) in ∂ui three steps: we introduce a new summation variable ι, take the partial derivative of the scalar coefficient field with respect to uι , and then add (dm uι ) to the list of fields in the sum. One can easily check that b ∂ = 0 for the operator so-defined satisfies conditions DO1 through DO6. And it is clear that ∇a ∂ui b ∂ all i. For when we (vacuously) represent any particular field in the indicated way, ∂ui0

∂ ∂ui0

b

=

n X i=1

i

α

∂ ∂ui

b

,

i

i

the coefficients α are constant (either 0 or 1), and so

∂α = 0 for all i and l. ∂ul

We call this derivative operator — the one identified in the proposition — the coordinate derivative operator canonically associated with (U, ϕ). Sometimes, when the there is no ambiguity about the n-chart b ∂ with which it is associated, the operator is written as ∂. So ∂a = 0 for all i. As we shall ∂ui see in the next section, all coordinate derivative operators are flat, i.e., their Riemann curvature fields vanish. Problem 1.7.3. Let ∇ be the coordinate derivative operator canonically associated with (U, ϕ) on the n-manifold M . Let ui be the coordinate maps on U determined by the chart. Further, let ∇′ be another

a derivative operator on U . We know (from proposition 1.7.3) that there is a smooth field Cbc on U such a that ∇′ = (∇, Cbc ). Show that if a Cbc

n X n X n X

=

i=1 j=1 k=1

then a smooth vector field ξ a =

n X i=1

i

ξ

∂ ∂ui

a

ijk

C

∂ ∂ui

a

(db uj ) (dc uk ),

on U is constant with respect to ∇′ (i.e., ∇′a ξ b = 0) iff

58


i

n X ijk k ∂ξ C ξ = j ∂u k=1

ijk

for all i and j. (The “Christoffel symbol” Γijk is often used to designate the coefficent field C .) Next, we make use of proposition 1.7.11 to prove a useful proposition about “position fields”. Proposition 1.7.12. Let ∇ be the coordinate derivative operator canonically associated with (U, ϕ) on the n-manifold M . Let ui be the coordinate maps on U determined by the chart, and let p be a point in

U . Then there exists a unique smooth vector field χa on U such that (1) ∇a χb = δab and (2) χa = 0 at p. Proof. (Existence) Consider the field χa defined by ∂ a . (1.7.11) u − u (p) χ = ∂ui i=1 Clearly it satisfies condition (2) since ui − ui (p) |p = ui (p) − ui (p) = 0. And it satisfies (1) because n X

a

∇a χ

b

=

n X i=1

i

i

b n X ∂ b ∂ i (∇a u ) = = δab . ∇a u − u (p) i ∂ui ∂u i=1 i

i

b ∂ are constant with respect to ∇; the ∂ui second equality follows from the fact that (all) derivative operators annihilate all constant scalar fields; (The first equality follows from the fact that the basis fields

and the third equality follows from equation (1.7.9).) (Uniqueness) Assume χ′a satisfies conditions (1) and (2) as well, and consider the difference field (χ′a − χa ). It is constant with respect to ∇ (since ∇a (χ′b − χb ) = δab − δab = 0), and it is the zero vector

at p. So it must be the zero vector everywhere, i.e., χ′a = χa .

We refer to χa as the position field relative to p (associated with the coordinate derivative ∇). In the last few paragraphs we have dealt with the derivative operator ∇ canonically associated with

an arbitrary n-chart (U, ϕ) on an arbitrary n-manifold M . Let us now consider the special case where M is the manifold Rn , (U, ϕ) is the (global) n-chart where U = Rn , and ϕ : U → Rn is the identity map. (So ui = xi ◦ ϕ = xi .) In this case, we get

∂f ∂xi

=

∂ (f ). ∂xi

from (1.7.5). Of course, we have already encountered the fields

(1.7.12)

∂ . They were the first examples of ∂xi

vector fields that we considered in section 1.3. (There we used (1.7.12) to characterize the fields.)

Many familiar textbook assertions about “differentials” fall out as consequences of the claims we have listed. For example, the equation df =

n X ∂f dxj j ∂x j=1

59


comes out in our notation as db f =

n X ∂f db xj , j ∂x j=1

and the latter is just an instance of (1.7.10). The coordinate derivative operator ∇ canonically associated with the coordinates x1 , ... , xn is defined

on the entire manifold Rn (because the coordinates are). So, too, the associated positions fields χa

(relative to particular points) are defined on the entire manifold. Note that we have encountered these position fields before as well. Suppose we take p to be the origin (i.e., suppose xi (p) = 0 for all i). Then, recalling (1.7.11), we have χa =

n X

xi

i=1

∂ ∂xi

a

.

In the case n = 2, the right side field is precisely what we called the “radius expansion” field in section 1.3. We can picture it as follows. Given any point q in R2 , there is a natural isomorphism between the vector space R2 and the tangent space to the manifold R2 at q defined by a a ∂ ∂ 2 + x . (x1 , x2 ) 7→ x1 |q ∂x1 ∂x2 |q → If we identify these two, then we can think of χa |q as just the “position vector” − oq that runs from the origin o to q. (See figure 1.7.1.)

χa |q q o

Figure 1.7.1: The position field χa on R2 (relative to point o).

60


1.8

Curvature

a In this section we introduce the Riemann curvature tensor field Rbcd and discuss its intuitive geometric

significance. We start with an existence claim. Lemma 1.8.1. Suppose ∇ is a derivative operator on the manifold M . Then there is a (unique) smooth a tensor field Rbcd on M such that for all smooth fields ξ b ,

a Rbcd ξ b = −2 ∇[c ∇d] ξ a .

(1.8.1)

Proof. Uniqueness is immediate since any two fields that satisfied this condition would agree in their a action on all vectors ξ b at all points. For existence, we introduce a field Rbcd and do so in such a way that 0

it is clear that it satisfies the required condition. Let p be any point in M and let ξ b be any vector at p. 0

0

a We define Rbcd ξ b by considering any smooth field ξ b on M that assumes the value ξ b at p and setting 0

a Rbcd ξb = −2 ∇[c ∇d] ξ a . It suffices to verify that the choice of the field ξ b plays no role. For this it suffices

to show that if η b is a smooth field on M that vanishes at p, then necessarily ∇[c ∇d] η b vanishes at p as

well. (For then we can apply this result, taking η b to be the difference between any two candidates for ξ b .) The usual argument works. Let λa be any smooth field on M . Then we have, by DO6, 0 = ∇[c ∇d] η a λa = ∇[c η a ∇d] λa + η a ∇[c ∇d] λa + ∇[c λ|a| ∇d] η a + λa ∇[c ∇d] η a .

(Note: In the third term of the final sum the vertical lines around the index indicate that it is not to be included in the anti-symmetrization.) Now the first and third terms in that sum cancel each other. And the second vanishes at p. So we have 0 = λa ∇[c ∇d] η a at p. But the field λa can be chosen so that it

assumes any particular value at p. So ∇[c ∇d] η a = 0 at p, as claimed.

a Rbcd is called the Riemann curvature tensor field (associated with ∇). It codes information about the

a degree to which the operators ∇c and ∇d fail to commute. Several basic properties of Rbcd are collected

in the next proposition.

Proposition 1.8.2. Suppose ∇ is a derivative operator on the manifold M . Then the curvature tensor a field Rbcd associated with ∇ satisfies the following conditions: ...ar (1) For all smooth tensor fields αba11...b on M , s

...ar a1 ...ar ...ar 2 ∇[c ∇d] αba11...b = αnb Rbn1 cd + ... + αba11...b Rbns cd s 2 ...bs s−1 n

a ...a

n

a1 ar 2 ...ar −αbna Rncd − ... − αb11...bsr−1 Rncd . 1 ...bs

a (2) Rb(cd) = 0. a (3) R[bcd] = 0. a (4) ∇[m R|b|cd] =0

(Bianchi’s identity).

61


Proof. Condition (1) is proved in the now familiar way using D06 and lemma 1.8.1. We proceed in n two steps. First, we show that 2 ∇[c ∇d] αb = αn Rbcd

for all fields αb on M . To do so, we consider an arbitary smooth field ξ on M , expand 0 = ∇[c ∇d] ξ αa , and invoke the lemma. Then we turn to a

a

...ar the general case. We contract αba11...b with s smooth contravariant vector fields and r smooth covariant s

vector fields, apply ∇[c ∇d] , expand, and then use our previous results. (2) follows immediately from lemma 1.8.1. For (3), notice that given any smooth scalar field α on M , we have, by (1), a Rbcd ∇a α = 2 ∇[c ∇d] ∇b α

and hence, by D06, a R[bcd] ∇a α = 2 ∇[c ∇d ∇b] α = 0.

Since any covariant vector at any point can be realized in the form ∇a α (recall lemma 1.7.2), it follows

a that R[bcd] = 0 everywhere.

The argument for (4) is just a bit more complicated. Given any smooth field αb on M , we have a a a 2 ∇r ∇[c ∇d] αb = ∇r (Rbcd αa ) = (∇r Rbcd )αa + Rbcd ∇r αa .

But we also have, by (1), n n 2 ∇[r ∇c] ∇d αb = Rdrc ∇n αb + Rbrc ∇d αn .

If we anti-symmetrize these two equations in (r, c, d), then we have 2 ∇[r ∇c ∇d] αb on the left side of both. So (equating their right sides),

a a n n ∇[r R|b|cd] αa + Rb[cd ∇r] αa = R[drc] ∇n αb + Rb[rc ∇d] αn .

The second term on the left here is equal to the second term on the right. So, by condition (3), we have

But αa is arbitrary, and so we have (4).

a ∇[r R|b|cd] αa = 0.

a Problem 1.8.1. Let ∇ and ∇′ be derivative operators on a manifold with ∇′m = (∇m , Cbc ), and let their

a respective curvature fields be Rbcd and R′abcd . Show that

a n a a . Cd]n + 2Cb[c R′abcd = Rbcd + 2∇[c Cd]b

(1.8.2)

Problem 1.8.2. Show that the exterior derivative operator d on any manifold satisfies d2 = 0, i.e., dn (dm αb1 ...bp ) = 0 for all smooth p-forms αb1 ...bp . (Hint: Make use of proposition 1.8.2. Notice also that λ[a...[b...c]...d] = λ[a...b...c...d] for all tensors λa...b...c...d .) Problem 1.8.3. Show that given any smooth field ξ a , and any derivative operator ∇ on a manifold, £ξ

m m commutes with ∇ (in its action on any tensor field) iff ∇a ∇b ξ m = Rbna ξ n . (Here, of course, Rbna is

62


the curvature field associated with ∇. If this conditions holds, we say that ξ a is an “affine collineation” ...ar m m with respect to ∇. Hint: First show that if Kab = Rbna ξ n − ∇a ∇b ξ m , then for all smooth fields αba11...b , s ...ar (£ξ ∇n − ∇n £ξ )αba11...b s

=

a1 ...ar a1 ...ar m m αm b2 ...bs Kn b1 + ... + αb1 ...bs−1 m Kn bs a ...a

a2 ...ar −αm Kna1m − ... − αb11...bsr−1 b1 ...bs

m

Knarm .)

Problem 1.8.4. Show that given any smooth field ξ a on a manifold, the operators £ξ and da commute in their action on all smooth p-forms. (Hint: Make use of the equation stated in the hint for problem 1.8.3.) It is not our purpose to attempt to develop systematically the theory of forms on a manifold, but we shall pause for one comment on the result stated in problem 1.8.2. Let αa1 ...an be a smooth n-form on a manifold M with n ≥ 1. We say that is closed if its exterior derivative vanishes. And we say that it

is exact if there is a (n − 1)-form on M of which it is the exterior derivative. (So, for example, the form

αab is closed if da αbc = 0, and it is exact if there is a smooth form βa such that αab = da βb .) It follows immediately from the problem that every exact form is closed. It turns out that the converse is true as well, at least locally, but the proof is non-trivial. We record the fact here for future reference. Proposition 1.8.3. Let αa1 ...an be a smooth closed n-form on the manifold M with n ≥ 1. Then,

for all p in M , there is an open set O containing p, and an (n − 1)-form βa1 ...an−1 on O such that αa1 ...an = da1 βa2 ...an .

Global assertions can also be made if M satisfies suitable conditions. If M is simply connected, for example, then all closed 1-forms are (globally) exact. And if M is contractible then, for all n ≥ 1, all

closed n-forms are (globally) exact. (See Spivak [57, volume 1] for proofs of the two claims. Proposition 1.8.3 is a consequence of the second, since all manifolds are locally contractible.) a Suppose M is a manifold with derivative operator ∇ and associated curvature field Rbcd . We say that

a ∇ is flat (or that M is flat relative to ∇) if Rbcd vanishes everywhere on M . The next proposition makes

clear the intuitive geometric significance of flatness.

Proposition 1.8.4. Let ∇ be a derivative operator on the manifold M . If parallel transport of vectors on M relative to ∇ is path independent, then ∇ is flat. Conversely, if ∇ is flat, then, at least locally

(i.e., within some open neighborhood of every point), parallel transport of vectors relative to ∇ is path independent. (If M is simply connected, the converse holds globally.)

Proof. First assume that parallel transport of vectors on M is path independent. Let p be any point in 0

0

M , and let ξ a be any vector at p. We extend ξ a to a smooth vector field ξ a on all of M by parallel 0

transporting ξ a (via any curve) to other points of M . The resulting field is constant in the sense that ∇a ξ b = 0 everywhere. (This follows from the fact that all directional derivatives of ξ b at all points

63


0

0

a a vanish.) Hence Rbcd ξ b = −2 ∇[c ∇d] ξ a = 0 at all points. In particular Rbcd ξ b = 0 at p. Since ξ b was

a arbitrary, we have Rbcd = 0 at p.

a Conversely, suppose that Rbcd vanishes on M . To show that parallel transport on M is, at least locally, 0

0

path independent, it will suffice to show that given any vector ξa at point p, there is an extension of ξa to a smooth field ξ a on some open set O containing p that is constant, i.e., ∇a ξ b = 0 everywhere in O. (For

then, given any point q ∈ O, and any curve γ from p to q whose image falls within O, parallel transport 0

0

a a of ξ a along γ must yield ξ|q .) To see that a vector field satisfying ∇a ξ b = 0 and ξ|p = ξ a does exist

locally, one writes out these two conditions in terms of local coordinates and generates a set of partial differential equations. These equations, it turns out, have a solution if a certain “integrability condition” a is satisfied. That condition is nothing but the equation Rbcd = 0 expressed in local coordinates. (For

further details, see, for example, Spivak [57], volume 2, chapter 4.) We know from proposition 1.7.11, that given any n-chart (U, ϕ) (with non-empty domain) on an b ∂ n-manifold, there is a unique derivative operator ∇ on U such that ∇a = 0 for all i. (Here ∂ui 1 n u , ..., u are the coordinate maps on U determined by (U, ϕ).) We called it the “coordinate derivative operator canonically associated with (U, ϕ)”. It follows immediately, of course, that a Rbcd

∂ ∂ui

b

= −2 ∇[c ∇d]

∂ ∂ui

a

=0

b b ∂ ∂ , ..., span the tangent ∂u1 ∂un space at every point. Thus we see that coordinate derivative operators canonically associated with local

a for all i. This, in turn, implies that Rbcd = 0, since the fields

charts are flat. The geometric significance of the curvature tensor field can also be explicated in terms of “geodesic deviation”. Suppose ξ a is a smooth vector field on the manifold M whose integral curves are geodesics with respect to ∇. (We shall say that ξ a is a geodesic field with respect to ∇.) Further suppose that λa

is a smooth field that satisfies £ξ λa = 0. Then we can think of the restriction of λa to an integral curve γ of ξ a as a field that connects γ to an “infinitesimally close” integral curve γ ′ . If we do, the second derivative field ξ n ∇n ξ m ∇m λa along γ represents the “relative acceleration” of γ ′ with respect to γ.

The following proposition shows how this field can be expressed in terms of the Riemann curvature field.

Proposition 1.8.5. Suppose ξ a is a geodesic field on the manifold M with respect to ∇. Further suppose

λa is a smooth field that satisfies £ξ λa = 0. Then

a ξ b λc ξ d . ξ n ∇n ξ m ∇m λa = Rbcd

(1.8.3)

Proof. We have ξ n ∇n ξ a = 0 (since ξ a is a geodesic field) and ξ n ∇n λa = λn ∇n ξ a (since £ξ λa = 0).

64


The rest is just a calculation. ξ n ∇n ξ m ∇m λa

= ξ n ∇n λm ∇m ξ a = ξ n ∇n λm ∇m ξ a + ξ n λm ∇n ∇m ξ a a = ξ n ∇n λm ∇m ξ a + ξ n λm ∇m ∇n ξ a + ξ n λm Rrmn ξr a ξr = ξ n ∇n λm ∇m ξ a +λm ∇m ξ n ∇n ξ a − λm ∇m ξ n ∇n ξ a +ξ n λm Rrmn

a = Rrmn ξ r λm ξ n .

a (The third equality follows from Rrmn ξ r = −2 ∇[m ∇n] ξ a . The final one follows from the fact that in the

sum before the equality sign, the second term is 0, and the first and third terms cancel each other.)

Proposition 1.8.6. Suppose ∇ is a derivative operator on the manifold M . Then ∇ is flat iff all geodesic

deviation on M (with respect to ∇) vanishes, i.e., given any smooth geodesic field ξ a , and any smooth field λa such that £ξ λa = 0, ξ n ∇n ξ m ∇m λa = 0.

Proof. The “only if” half follows immediately from proposition 1.8.5. So suppose that all geodesic 0

0

0

0

0

a deviation vanishes. Then, given any vectors ξa and λa at a point p, it must be the case that Rbcd ξb λc ξd =

0. (We can always choose field ξ a and λa on an open set containing p such that ξ a is a geodesic field, 0

0

£ξ λa = 0, and ξ a and λa assume the values ξ a and λ a at p respectively.) Equivalently, it must be 0

0

0

a the case that Rbcd ξ b ξ d = 0 for all vectors ξ b at p. Our conclusion now follows by the symmetries of

the Riemann tensor recorded as conditions (2) and (3) in proposition 1.8.2. By (2), first, it follows that 0

0

0

a Rbcd ξ b ξ c = 0 for all vectors ξ b at p. Hence, by proposition 1.4.3, a R(bc)d =0

(1.8.4)

at p. Next, by (2) and (3), we have (everywhere) a a a Rbcd + Rdbc + Rcdb = 0.

But (2) and (1.8.4) jointly imply a a a Rbcd = Rdbc = Rcdb a at p. So Rbcd = 0 at our arbitrary point p.

Equation (1.8.3) is called the equation of geodesic deviation. Notice that it must be the second derivative field ξ n ∇n ξ m ∇m λa that enters the equation, and not the first derivative field ξ n ∇n λa . The latter is unconstrained by the curvature of the manifold. It can assume any value at a point.


1.9

65

Metrics

A (semi-Riemannian) metric on a manifold M is a smooth field gab on M that is symmetric and invertible, i.e., there exists an (inverse) field g bc on M such that gab g bc = δac . It is easy to check that the inverse field g bc of a metric gab is symmetric and unique. It is symmetric since b mc g cb = g nb δnc = g nb (gnm g mc ) = (gmn g nb ) g mc = δm g = g bc .

(Here we use the symmetry of gnm for the third equality.) It is unique because if g ′bc is also an inverse field, then c g ′bc = g ′nc δnb = g ′nc (gnm g mb ) = (gmn g ′nc ) g mb = δm g mb = g cb = g bc .

(Here again we use the symmetry of gnm for the third equality; and we use the symmetry of g cb for the final equality.) One can also check that the inverse field g bc of a metric gab is smooth. This follows, essentially, because given any invertible square matrix A (over R), the components of the inverse matrix A−1 depend smoothly on the components of A. The requirement that a metric be invertible can be given a second formulation. Indeed, given any field gab on the manifold M (not necessarily symmetric and not necessarily smooth), the following conditions are equivalent. (1) There is a tensor field g bc on M such that gab g bc = δac . (2) For all p in M , and all vectors ξ a at p, if gab ξ a = 0, then ξ a = 0. (When the conditions obtain, we say that gab is non-degenerate.) To see this, assume first that (1) holds. Then given any vector ξ a at any point p, if gab ξ a = 0, it follows that ξ c = δac ξ a = g bc gab ξ a = 0. Conversely, suppose that (2) holds. Then at any point p, the map from (Mp )a to (Mp )b defined by ξ a 7→ gab ξ a is an injective linear map. Since (Mp )a and (Mp )b have the same dimension, it must be

surjective as well. So the map must have an inverse g bc defined by g bc (gab ξ a ) = ξ c or g bc gab = δac .

In the presence of a metric gab , it is customary to adopt a notation convention for “lowering and raising indices”. Consider first the case of vectors. Given a contravariant vector ξ a at some point, we write gab ξ a as ξb ; and given a covariant vector ηb , we write g bc ηb as η c . The notation is evidently consistent in the sense that first lowering and then raising the index of a vector (or vice versa) leaves the vector intact. One would like to extend this notational convention to tensors with more complex index structure. But now one confronts a problem. (It was mentioned in passing in section 1.4.) Given a tensor αab c at a mab point, for example, how should we write g mc αab ? Or as αamb ? Or as αabm ? In general, c ? As α

these three tensors will not be equal. To get around the problem, we introduce a new convention. In any context where we may want to lower or raise indices, we shall write indices, whether contravariant or covariant, in a particular sequence. So, for example, we shall write αabc or αac b or αcab . (These

66


tensors may be equal — they belong to the same vector space — but they need not be.) Clearly this convention solves our problem. We write g mc αabc as αabm ; g mc αac b as αamb ; and so forth. No ambiguity arises. (And it is still the case that if we first lower an index on a tensor and then raise it (or vice versa), the result is to leave the tensor intact.) We claimed in the preceding paragraph that the tensors αabc and αac b (at some point) need not be equal. Here is an example. (It is just a variant of the one used in section 1.4 to show that the tensors 1

2

n

αab and αba need not be equal.) Suppose ξ a , ξ a , ... , ξ a is a basis for the tangent space at a point p. i

j

k

i

j

i

j

k

Further suppose αabc = ξa ξb ξc at the point. Then αacb = ξa ξc ξb there. Hence, lowering indices, we have i

j

k

k

αabc = ξ a ξ b ξ c but αac b = ξ a ξ c ξ b at p. These two will not be equal unless j = k. We have reserved special notation for two tensor fields: the index substiution field δba and the Riemann a curvature field Rbcd (associated with some derivative operator). Our convention will be to write these as

δ ab and Rabcd , i.e., with contravariant indices before covariant ones. As it turns out, the order does not matter in the case of the first since δ ab = δba . (It does matter with the second.) To verify the equality, it suffices to observe that the two fields have the same action on an arbitrary field αb : δba αb = (gbn g am δ nm )αb = gbn g an αb = gbn g na αb = δ ab αb . Similarly we can verify (if we are raising and lowering indices with gab ) that δ ab = g ab and δab = gab . (We shall take these different equalities for granted in what follows.) Now suppose gab is a metric on the n-dimensional manifold M and p is a point in M . Then there 1

2

n

exists an m, with 0 ≤ m ≤ n, and a basis ξ a , ξ a , ... , ξ a for the tangent space at p such that i

i

i

i

gab ξ a ξ b gab ξ a ξ b i

=

+1

if

1 ≤ i ≤ m,

=

−1

if

m < i ≤ n,

=

0

if

i 6= j.

j

a b

gab ξ ξ

Such a basis is called orthonormal. Orthonormal bases at p are not unique, but all have the same associated number m. We call the pair (m, n−m) the signature of gab at p. (The existence of orthonormal bases and the invariance of the associated number m are basic facts of linear algebraic life. See, for example, Lang [36].) A simple continuity argument shows that any connected manifold must have the same signature at each point. In what follows we shall restrict attention to connected manifolds and refer simply to the “signature of gab ”. A metric with signature (n, 0) is said to be positive definite. With signature (0, n), it is said to be negative definite. With any other signature it is said to be indefinite. One case will be of special interest to us later. A Lorentzian metric is a metric with signature (1, n − 1). The mathematics of relativity

theory is, to some degree, just a chapter in the theory of four-dimensional manifolds with Lorentzian metrics. 1

2

n

Suppose gab has signature (m, n − m), and ξ a , ξ a , ... , ξ a is an orthonormal basis at a point. Further, P P i i i i suppose µa and ν a are vectors there. If µ a = ni=1 µ ξ a and ν a = ni=1 ν ξ a , then it follows from the

67


linearity of gab that 1 1

mm

m+1 m+1

gab µa ν b = µ ν + ... + µ ν − µ

ν

n n

− ... − µ ν.

(1.9.1)

In the special case where the metric is positive definite, this comes to 1 1

nn

gab µa ν b = µ ν + ... + µ ν.

(1.9.2)

And where it is Lorentzian, 1 1

2 2

nn

gab µa ν b = µ ν − µ ν − ... − µ ν.

(1.9.3)

So far we have introduced metrics and derivative operators as independent objects. But, in a quite natural sense, a metric determines a unique derivative operator. Suppose gab and ∇ are both defined on the manifold M . Further suppose γ : I → M is a smooth

curve on M with tangent field ξ a and λa is a smooth field on γ. Both ∇ and gab determine a criterion of

“constancy” for λa . λa is constant with respect to ∇ if ξ n ∇n λa = 0. λa is constant with respect to gab if gab λa λb is constant along γ, i.e., if ξ n ∇n gab λa λb = 0. It seems natural to consider pairs gab and

∇ for which the first condition of constancy implies the second.

Let us say that ∇ is compatible with gab if, for all γ and λa as above, λa is constant with respect to

gab whenever it is constant with respect to ∇. The next lemma gives the condition a more economical formulation.

Lemma 1.9.1. Suppose ∇ is a derivative operator, and gab is a metric, on the manifold M . Then ∇ is compatible with gab iff ∇a gbc = 0.

Proof. Suppose γ is an arbitrary smooth curve with tangent field ξ a and λa is an arbitrary smooth field on γ satisfying ξ n ∇n λa = 0. Then ξ n ∇n gab λa λb

= gab λa ξ n ∇n λb + gab λb ξ n ∇n λa + λa λb ξ n ∇n gab | {z } | {z } = 0

a

= 0

b n

= λ λ ξ ∇n gab .

Suppose first that ∇n gab = 0. Then it follows immediately that ξ n ∇n gab λa λb = 0. So ∇ is compatible

with gab . Suppose next that ∇ is compatible with gab . Then for all choices of γ and λa (satisfying

ξ n ∇n λa = 0), we have λa λb ξ n ∇n gab = 0. Since the choice of λa (at any particular point) is arbitrary

and gab is symmetric, it follows (by proposition 1.4.3) that ξ n ∇n gab = 0. But this must be true for arbitrary ξ a (at any particular point), and so we have ∇n gab = 0.

Note that the condition of compatibility is also equivalent to ∇a g bc = 0. To see this, recall (problem

1.7.1) that ∇a δ mn = 0. Hence, 0

= g bn ∇a δ cn = g bn ∇a gnr g rc = g bn gnr ∇a g rc + g bn g rc ∇a gnr

= δ br ∇a g rc + g bn g rc ∇a gnr = ∇a g bc + g bn g rc ∇a gnr .


68

So if ∇a gbc = 0, it follows immediately that ∇a g bc = 0. Conversely, if ∇a g bc = 0, then g bn g rc ∇a gnr =

0. And therefore,

0 = gpb gsc g bn g rc ∇a gnr = δ np δ rs ∇a gnr = ∇a gps . The basic fact about compatible derivative operators is the following. Proposition 1.9.2. Suppose gab is a metric on the manifold M . Then there is a unique derivative operator on M that is compatible with gab . Proof. To prove that M admits any derivative operator at all is a bit involved, and we skip the argument. (See Geroch [23]. It turns out that if a manifold admits a metric, then it necessarily satisfies the countable cover condition (M5) that we considered in section 1.1. And the latter, as noted in proposition 1.7.1, guarantees the existence of a derivative operator.) We do prove that if M admits a derivative operator ∇, then it admits exactly one ∇′ that is compatible with gab . Every derivative operator ∇′ on M can be realized as ∇′ = (∇, C abc ), where C abc is a smooth,

symmetric field on M . Now

∇′a gbc = ∇a gbc + gnc C nab + gbn C nac = ∇a gbc + Ccab + Cbac . So ∇′ will be compatible with gab (i.e., ∇′a gbc = 0) iff ∇a gbc = −Ccab − Cbac .

(1.9.4)

Thus it suffices for us to prove that there exists a unique smooth, symmetric field C abc on M satisfying (1.9.4). To do so, we write equation (1.9.4) twice more after permuting the indices: ∇c gab

= −Cbca − Cacb ,

∇b gac

= −Ccba − Cabc .

If we subtract these two from the first equation, and use the fact that Cabc is symmetric in (b, c), we get Cabc = and, therefore, C abc =

1 ∇a gbc − ∇b gac − ∇c gab , 2

1 an ∇n gbc − ∇b gnc − ∇c gnb . g 2

(1.9.5)

(1.9.6)

This establishes uniqueness. But clearly the field C abc defined by (1.9.6) is smooth, symmetric, and satisfies (1.9.4). So we have existence as well. In the case of positive definite metrics, there is another way to capture the significance of compatibility of derivative operators with metrics. Suppose the metric gab on M is positive definite, and γ : [s1 , s2 ] → M

69


is a smooth curve on M .3 We associate with γ a length Z s2 1 gab ξ a ξ b 2 ds, |γ| = s1

where ξ a is the tangent field to γ. This assigned length is invariant under reparametrization. For suppose σ : [t1 , t2 ] → [s1 , s2 ] is a diffeomorphism we shall write s = σ(t) and ξ ′a is the tangent field of ds γ ′ = γ ◦ σ : [t1 , t2 ] → M . Then ξ ′a = ξ a . (Recall equation (1.3.1) in the proof of proposition 1.3.2.) dt We may as well require that the reparametrization preserve the orientation of the original curve, i.e., ds > 0 everywhere. (Only small changes are require that σ(t1 ) = s1 and σ(t2 ) = s2 . In this case, dt ds 0 such that, for

all s1 , s2 ∈ I with s1 ≤ s ≤ s2 , if s2 − s1 < ε and if γ ′ = γ|[s1 ,s2 ] (the restriction of γ to [s1 , s2 ]), then |γ ′ | = d γ(s1 ), γ(s2 ) . Intuitively, minimal curves are “locally shortest curves”. Certainly they need not

be “shortest curves” outright. (Consider, for example, two points on the “equator” of a two-sphere that

are not antipodal to one another. An equatorial curve running from one to the other the “long way” qualifies as a minimal curve.) One can characterize the unique derivative operator compatible with a positive definite metric gab in terms of the latter’s associated minimal curves. But in doing so one has to pay attention to parametrization. Let us say that a smooth curve γ : I → M with tangent field ξ a is parametrized by arc length if for

all ξ a , gab ξ a ξ b = 1. In this case, if I = [s1 , s2 ], then Z s2 Z 1 gab ξ a ξ b 2 ds = |γ| = s1

3 Officially

s2

s1

1 ds = s2 − s1 .

(in section 1.2), we have taken a “smooth curve on M ” to be a smooth map of the form γ : I → M where

I is an open (possibly infinite or half infinite) interval in R. Let us now agree to extend the definition and allow for the

possibility that the interval I is not open. In this case, we take γ to be smooth if there is an open interval I ′ ⊆ R, with

I ⊆ I ′ , and a smooth map γ ′ : I ′ → M , such that γ ′ (s) = γ(s) for all s ∈ I. And in this case, of course, we obtain the

“tangent field of γ” by restricting that of γ ′ to I. Furthermore, if σ : I ′ → I is a bijection betweeen (not necessarily open)

intervals in R, we understand it to be a diffeomorphism if σ and σ−1 are both smooth in the sense just given.


70

(Any non-trivial smooth curve can always be reparametrized by arc length.) Our characterization theorem is the following. Proposition 1.9.3. Suppose gab is a positive definite metric on the manifold M and ∇ is a derivative

operator on M . Then ∇ is compatible with gab iff for all smooth curves γ parametrized by arc length, γ is a geodesic with respect to ∇ iff it is minimal with respect to gab .

Note that the proposition would be false if the qualification “parametrized by arc length” were dropped. The class of minimal curves is invariant under reparametrization. The class of geodesics (determined by a derivative operator) is not. We skip the proof of proposition of 1.9.3, which involves ideas from the calculus of variations. And we assert, without further discussion at this stage, that more complicated versions of the theorem are available when the metric gab under consideration is not positive definite. (We shall later consider the Lorentzian case.) We have already demonstrated (proposition 1.8.2) that the Riemann tensor field associated with any derivative operator exhibits several index symmetries. When the derivative operator is determined by a metric, yet further symmetries are present. Proposition 1.9.4. Suppose gab is a metric on a manifold M , ∇ is the derivative operator on M

compatible with gab , and Rabcd is associated with ∇. Then Rabcd (= gam Rmbcd ) satisfies the following conditions.

(1) Rab(cd) = 0. (2) Ra[bcd] = 0. (3) R(ab)cd = 0. (4) Rabcd = Rcdab . Proof. (1) and (2) follow directly from clauses (2) and (3) of proposition 1.8.2. And by clause (1) of that proposition, we have, since ∇a gbc = 0, 0 = 2 ∇[c ∇d] gab = gnb Rnacd + gan Rnbcd = Rbacd + Rabcd . That gives us (3). So it will suffice for us to show that clauses (1) – (3) jointly imply (4). Note first that 0

= Rabcd + Radbc + Racdb = Rabcd − Rdabc − Racbd .

(The first equality follows from (2), and the second from (1) and (3).) So anti-symmetrization over (a, b, c) yields 0 = R[abc]d − Rd[abc] − R[acb]d .

71


Rabcd

Rcbda

Rcabd

Rbdca Radbc

Rcdab Figure 1.9.1: Symmetries of the Riemann tensor field Rabcd .

The second term is 0 by clause (2) again, and R[abc]d = −R[acb]d . So we have an intermediate result: R[abc]d = 0.

(1.9.7)

Now consider the octahedron in figure 1.9.1. Using (1) – (3) and (1.9.7), one can easily verify that the sum of the terms corresponding to each triangular face vanishes. For example, the shaded face determines the sum Rabcd + Rbdca + Radbc = −Rabdc − Rbdac − Rdabc = −3 R[abd]c = 0. So if we add the sums corresponding to the four upper faces, and subtract the sums corresponding to the four lower faces, we get (since “equatorial” terms cancel), 4 Rabcd − 4 Rcdab = 0. This gives us (4). We say that two metrics gab and g ′ab on a manifold M are projectively equivalent if their respective associated derivative operators are projectively equivalent, i.e., if their associated derivative operators admit the same geodesics up to reparametrization. (Recall our discussion in section 1.7.) In contrast, we say that they are conformally equivalent if there is a map Ω : M → R such that g ′ab = Ω2 gab .


72

Ω is called a conformal factor. (If such a map exists, it must be smooth and non-vanishing since both gab and g ′ab are.) Notice that if gab and g ′ab are conformally equivalent, then, given any point p, and any vectors ξ a and η a at p, they agree on the ratio of their assignments to the two, i.e., gab ξ a ξ b g ′ab ξ a ξ a = g ′ab η a η b gab η a η b (if the denominators are non-zero). If two metrics are conformally equivalent with conformal factor Ω, then the connecting tensor field C abc that links their associated derivative operators can be expressed as a function of Ω. Proposition 1.9.5. If gab and g ′ab = Ω2 gab are metrics on the manifold M , and ∇′ = ∇, C abc , then i 1 h C abc = − 2 δ ab ∇c Ω2 + δ ac ∇b Ω2 − gbc g ar ∇r Ω2 . (1.9.8) 2Ω Proof. Since ∇′ is compatible with g ′ab , it follows that i 1h g ′dr C rbc = ∇d g ′bc − ∇b g ′dc − ∇c g ′db . 2

(Recall (1.9.5) in the proof of proposition 1.9.2.) If we substitute Ω2 gab for g ′ab and use the fact that

∇ is compatible with gab , this gives us Ω2 gdr C rbc = Contracting both sides with g ad yields Ω2 C abc = as claimed.

i 1h gbc ∇d Ω2 − gdc ∇b Ω2 − gdb ∇c Ω2 . 2

i 1h gbc g ad ∇d Ω2 − δ ac ∇b Ω2 − δ ab ∇c Ω2 , 2

The next proposition asserts that if metrics are both projectively and conformally equivalent, then they can differ by at most a multiplicative constant. (The converse implication is immediate.) The result will later (in section 2.1) be of crucial importance in our discussion of the physical signficance of the spacetime metric. Proposition 1.9.6. Suppose the hypotheses of proposition 1.9.5 obtain and, in addition, gab and g ′ab are projectively equivalent. Further suppose that the dimension of M is at least 2. Then Ω is constant on M . Proof. Let the dimension of M be n ≥ 2. We know that C abc must satisfy equation (1.9.8). But by

proposition 1.7.10, we also have

C abc = δ ab ϕc + δ ac ϕb

(1.9.9)

for some smooth field ϕc . The proof proceeds by playing off equations (1.9.8) and (1.9.9) against each other. Contracting the two equations (and using the fact that δ aa = n), we get i n 1 h C aba = − 2 ∇b Ω2 + n ∇b Ω2 − ∇b Ω2 = − 2 ∇b Ω2 , 2Ω 2Ω C aba = ϕb + n ϕb = (n + 1)ϕb .

73


So −

1 n+1 ∇b Ω2 = ϕb . 2 2Ω n

(1.9.10)

Substituting into (1.9.8), this yields C abc =

i n + 1h a δ b ϕc + δ ac ϕb − gbc g ar ϕr . n

Comparing this expression for C abc with (1.9.9), we get

δ ab ϕc + δ ac ϕb = (n + 1)gbc g ar ϕr . If we contract both sides with g bc , we are left with ϕa + ϕa = (n + 1) n ϕa . Hence, since n ≥ 2, ϕa = 0. So ∇b Ω2 = 0, by equation (1.9.10). Note that in one-dimensional manifolds, all metrics are projectively equivalent. (All smooth curves are geodesics up to reparametrization with respect to all derivative operators.) For this reason the proposition fails if n = 1. In the case where a derivative operator ∇ is determined by a metric gab , the Riemann tensor field

Rabcd

associated with the former admits an instructive decomposition. Consider first the Ricci tensor

field Rab and scalar curvature field R defined by: Rab

=

Rcabc

R

=

Raa (= g ar Rra ).

The first is symmetric since, by (1), (3), and (4) of proposition 1.9.4, Rab = g cd Rdabc = g cd Rcbad = Rba . It also follows from the symmetries listed in proposition 1.9.4 that these are, up to sign, the only fields that can be obtained by contraction from Rabcd . (Contraction on any two indices yields either the zero field or ±Rab and, therefore, contraction on all four indices (two at a time) yields either the zero field or

±R.)

Problem 1.9.1. Let ∇ be a derivative operator on a manifold M compatible with the metric gab . Use

the Bianchi identity (in proposition 1.8.2) to show that ∇a Rab −

1 ab g R = 0. 2

(This equation will figure later in our discussion of Einstein’s equation.)

74


The Weyl (or conformal ) tensor field Cabcd is defined by Cabcd = Rabcd −

2 2 ga[d Rc]b + gb[c Rd]a − R ga[c gd]b n−2 (n − 1)(n − 2)

(1.9.11)

(if the dimension n of the underlying manifold is at least 3). The second and third terms on the right side exhibit symmetries (1) – (4) from proposition 1.9.4. Therefore, Cabcd does so as well. Furthermore, as is easily checked, C abca = 0. So all contractions of Cabcd vanish. Thus (1.9.11) provides a decomposition of Rabcd in terms of Rab , R, and that part of Rabcd whose contractions all vanish. Later we shall see that Einstein’s equation in relativity theory correlates Rab and R with the presence of mass-energy but does not constrain Cabcd . So, in a sense, the Weyl field is that part of the full Riemann curvature field that is left free by the dynamical constraints of the theory. It turns out that the Weyl field is conformally invariant, i.e., we have the following basic result. Proposition 1.9.7. Let gab and g ′ab = Ω2 gab be metrics on a manifold with respective Weyl fields Cabcd and C ′abcd . Then C ′a bcd = C a bcd . One can prove this with a laborious but straightforward calculation using problem 1.8.1 and proposition 1.9.5. (See Wald [60, pp. 446-47].) We have said that a metric gab is flat if its associated Riemann tensor field Rabcd vanishes everywhere. In parallel, we say that it is conformally flat if its Weyl tensor field Cabcd vanishes everywhere. It follows immediately from proposition 1.9.7 (and the definition of Cabcd ) that if a metric is conformally equivalent to a flat metric, then it is conformally flat. It turns out that the converse is true as well in manifolds of dimension at least 4. (In dimension 3, Cabcd vanishes automatically.) Our next topic is “isometries” and “Killing vector fields”. Given two manifolds with a metric, (M, gab ) and (M ′ , g ′ab ), we say that a smooth map ϕ : M → M ′

is an isometry if ϕ∗ (g ′ab ) = gab . (Recall our

discussion of “pull-back maps” in section 1.5.) This condition captures the requirement that ϕ preserve inner products. To see this, consider any point p in M and any two vectors ξ a and ρa at p. The two have an inner product gab |p ξ a ρb at p. The push-forward map (ϕp )∗ carries them to vectors (ϕp )∗ (ξ a ) and (ϕp )∗ (ρa ) at ϕ(p), whose inner product there is g ′ab |ϕ(p) (ϕp )∗ (ξ a ) (ϕp )∗ (ρb ) . In general, there

is no reason why these two inner products should be equal. But they will be if ϕ∗ (g ′ab ) = gab , for then gab |p ξ a ρb = ϕ∗ (g ′ab ) |p ξ a ρb = g ′ab |ϕ(p) (ϕp )∗ (ξ a ) (ϕp )∗ (ρb ) .

The second equality is just an instance of the condition (equation 1.5.2) that defines ϕ∗ (g ′ab ). Now suppose λa is a smooth (not necessarily complete) vector field on M . We say that λa is a Killing field (with respect to gab ) if £λ gab = 0 or, equivalently, if it satisfies “Killing’s equation” ∇(a λb) = 0.

(1.9.12)

75


(Here ∇ is understood to be the derivative operator on M compatible with gab .) Equivalence here follows

from proposition 1.7.4:

£λ gab = λn ∇n gab + gnb ∇a λn + gan ∇b λn = ∇a λb + ∇b λa . Note that λa is a Killing field ⇐⇒ the (local) flow maps determined by λa are isometries. This assertion is just a special case of proposition 1.6.6, and it explains the classical description of Killing fields as “infinitesimal isometries”. The following proposition is useful when one undertakes to find or classify Killing fields. Proposition 1.9.8. Let gab be a metric on the manifold M with associated derivative operator ∇. Further, let λa be a Killing field on M (with respect to gab ). Then

∇a ∇b λc = −Rm abc λm . Proof. Given any smooth field λa on M , we have 2 ∇[a ∇b] λc

= Rmcab λm ,

2 ∇[c ∇a] λb

= Rmbca λm ,

2 ∇[b ∇c] λa

= Rmabc λm .

If we subtract the third equation from the sum of the first two, and then use the fact that ∇(r λs) = 0,

we get

2 ∇a ∇b λc

=

(Rmcab + Rmbca − Rmabc )λm

=

3 Rm[abc] λm − 2 Rmabc λm .

But Rm[abc] = 0, and so our claim follows. In the following problems, assume that gab is a metric on a manifold M and ∇ is its associated

derivative operator.

Problem 1.9.2. Let ξ a be a smooth vector field on M . Show that £ξ g ab = 0 ⇐⇒ £ξ gab = 0. Problem 1.9.3. Show that Killing fields on M with respect to gab are affine collineations with respect to ∇. (Recall problem 1.8.3.) Problem 1.9.4. Show that if ξ a is a Killing field on M with respect to gab , then the Lie derivative operator £ξ annihilates the fields Rabcd , Rab , and R (determined by gab ).

76


Problem 1.9.5. Show that if ξ a and η a are Killing fields on M (with respect to gab ), and k is a real number, then (ξ a + η a ), (k ξ a ), and the commutator [ξ, η]a = £ξ η a are all Killing fields as well. (Thus, the set of Killing fields has the structure of a Lie algebra.) Problem 1.9.6. Let η a be a Killing field on M with respect to gab . (i) Let γ be a geodesic with tangent field ξ a . Show that the function E = ξ a ηa is constant on γ. (ii) Let T ab be a smooth tensor field that is symmetric and divergence free (i.e., ∇a T ab = 0), and let J a be the field T ab ηb . Show that ∇a J a = 0. (Both of these assertions will be important later in connection with our discussion of conservation principles.)

Problem 1.9.7. A smooth field η a on M is said to be a “conformal Killing field” (with respect to gab ) if £η Ω2 gab = 0 for some smooth scalar field Ω. Show that if η a is a conformal Killing field, and M has dimension n, then

∇(a ηb) =

1 (∇c η c ) gab . n

The set of Killing fields on a manifold with a metric has a natural vector space structure (problem 1.9.5). It turns out that if n is the dimension of the manifold and d is the dimension of this vector space, then 0 ≤ d ≤

1 2

n (n + 1). We will not prove this inequality but will show that “n-dimensional Euclidean

space” does, in fact, admit

1 2

n (n + 1) linearly independent Killing fields.

Let ∇ be the flat derivative operator on the manifold Rn (with n ≥ 1) canonically associated with

1 n the (globally defined) projection coordinate our discussion toward the end of maps a x , ..., x . (Recall a ∂ ∂ section 1.7.) We know that the basis fields , ..., and co-basis fields (da x1 ), ..., (da xn ) are ∂x1 ∂xn all constant with respect to ∇. We take the Euclidean metric on Rn to be the field

gab = (da x1 )(db x1 ) + ... + (da xn )(db xn )

(1.9.13)

and take n-dimensional Euclidean space to be the pair (Rn , gab ). It follows that gab

∂ ∂xj

a

∂ ∂xk

for all j and k. Thus the fields

b

=

∂ ∂x1

a b X n ∂ ∂ i (da xi ) δij δik = δjk (d x ) = b ∂xj ∂xk i=1 i=1

n X

a

, ...,

∂ ∂xn

a

form an orthonormal basis for gab at every point, n ∂ ∂ and the signature of gab is (n, 0). It also follows that = g = (da xi ) for all i. (This an ∂xi a ∂xi does not hold in general. For example, as we shall see later, when we raise and lower indices with the ∂ n = −(da xi ) for some choices of i.) Hence Minkowski metric on R , ∂xi a a ∂ ∂ a i an i an = (d x ) = g (dn x ) = g ∂xi n ∂xi for all i and, therefore, the inverse metric field g ab can be expressed in the form g ab =

∂ ∂x1

a

∂ ∂x1

b

+ ... +

∂ ∂xn

a

∂ ∂xn

b

.

(1.9.14)

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Note also that ∇a gbc = 0, i.e., that ∇ is the unique derivative operator compatible with gbc . (This follows immediately since the scalar coefficient fields on the right side of (1.9.13) are all constant.)

Now we proceed to find all Killing fields in n-dimensional Euclidean space. Doing so is easy given the machinery we have developed. Proposition 1.9.9. Let ξ a be a Killing field in n-dimensional Euclidean space (Rn , gab ) (with n ≥ 1), let ∇ be the flat derivative operator on Rn canonically associated with the projection coordinate maps

x1 , ..., xn (which is compatible with gab ), let p be any point in Rn , and let χa be the position field on Rn

determined relative to p and ∇. (Recall proposition 1.7.12.) Then the following both hold. (1) There exist a unique constant, anti-symmetric field Fab and a unique constant field k a such that ξb = χa Fab + kb .

(1.9.15)

(Here, of course, “constant” means constant with respect to ∇.) (2) The vector space of Killing fields in (Rn , gab ) has dimension

1 2

n (n + 1).

Proof. (1) (Existence) Consider the fields Fab = ∇a ξb , and kb = ξb − χa Fab . Since ξ a is a Killing field,

∇(a ξb) = 0. So Fab is anti-symmetric. Clearly, the two fields satisfy equation 1.9.15. So what we need

to show is that they are both constant with respect to ∇. Fab is, since ∇n Fab = ∇n ∇a ξb = −Rmnab ξm =

0. (The second equality follows from proposition 1.9.8, and the third from the fact that ∇ is flat.) Furthermore, kb is constant, since

∇n kb = ∇n ξb − ∇n (χa Fab ) = Fnb − Fab ∇n χa = Fnb − Fab δ an = Fnb − Fnb = 0. (For the second equality we use the fact that ∇n Fab = 0, and for the third that ∇n χa = δ an .) (Uniqueness) Assume that the fields F ′ab and k ′a also satisfy the stated conditions. It follows that Fab

=

Fnb δ n a = Fnb ∇a χn = ∇a (χn Fnb + kb ) = ∇a ξb = ∇a (χn F ′nb + k ′b )

=

F ′nb ∇a χn = F ′nb δ n a = F ′ab

and, therefore, kb = k ′b . (2) Let d be the dimension of the vector space of Killing fields in (Rn , gab ). It follows from part (1) that d is of the form d = d1 + d2 , where d1 is the dimension of the vector space of constant, anti-symmetric n a fields Fabon (R a , gab ), and da2 is the dimension of the vector space of constant fields k on the manifold. n(n − 1) ∂ ∂ , ..., form a basis for the latter. So d2 = n. We claim that d1 = . This Clearly, ∂x1 ∂xn 2 n(n − 1) n(n + 1) will suffice, of course, for then d = n + = . To verify the claim, consider the expansion 2 2 of any constant, anti-symmetric field Fab in terms of the co-basis fields (da x1 ), ..., (da xn ). The coefficient

fields are all constant (since Fab is). So they determine an n × n anti-symmetric (real) matrix. (The ij th

entry is the coefficient of (da xi )(db xj ) in the expansion.) Thus the problem reduces to that of determining the dimension of the vector space of all n × n anti-symmetric real matrices. Since all numbers on the

78


diagonal must be 0, and the ij th and jith entries must sum to 0, the number of independent entries is n(n − 1) just the number of ordered pairs (i, j) where 1 ≤ i < j ≤ n. And this number is certainly . So 2 we are done.

p

Figure 1.9.2: Killing fields in the Euclidean plane. 3 × 2 2 a linearly independent aKillingfields. a Here the space of constant vector fields k is two-dimensional and is ∂ ∂ and . The space of constant, anti-symmetric fields Fab is one-dimensional generated by ∂x1 ∂x2 and is generated by Consider, for example, the case of two-dimensional Euclidean space where there should be 3 =

Fab = (da x1 )(db x2 ) − (da x2 )(db x1 ). So the full vector space of Killing fields is generated by the three fields 1

ξb

=

∂ ∂x1

b b

b

=

ξb

=

χa Fa b =

2

ξ 3

∂ ∂x2

∂ b ∂ b 2 2 x1 − x1 (p) − x − x (p) . ∂x2 ∂x1

The expression for the third is easily derived using our expression for Fab and (1.7.11) (in the case where ui = xi ): ξa

= = = =

χa Fan g nb a a ∂ ∂ 2 2 (x1 − x1 (p)) + (x − x (p)) (da x1 )(dn x2 ) − (da x2 )(dn x1 ) g nb 1 2 ∂x ∂x a a " b b # ∂ ∂ ∂ ∂ (da x1 ) + (x2 − x2 (p)) − (da x2 ) (x1 − x1 (p)) ∂x1 ∂x2 ∂x2 ∂x1 b b ∂ ∂ 2 2 (x1 − x1 (p)) − (x − x (p)) . ∂x2 ∂x1

79


The first two are the “infinitesimal generators” of horizontal and vertical translations. (See figure 1.9.2.) The third is the generator of counterclockwise rotations centered at p. If p = (0, 0), the third b b ∂ ∂ 2 − x that we have already encountered in section 1.3. reduces to the field x1 ∂x2 ∂x1 Finally, we briefly consider “manifolds of constant curvature”, a topic that will arise when we discuss Friedmann spacetimes in section 2.11. We say that a manifold with metric (M, gab ) has constant curvature κ at a point in M if Rabcd = κ (gad gbc − gac gbd )

(1.9.16)

holds there. (And, of course, we say that is has constant curvature at a point if it has constant curvature κ there for some κ.) Note that it is “possible” for (1.9.16) to hold only because the field

gabcd =

(gad gbc − gac gbd ) exhibits the same index symmetries as Rabcd (recall proposition 1.9.4): R(ab)cd = 0

g(ab)cd = 0,

(1.9.17)

Rab(cd) = 0

gab(cd) = 0,

(1.9.18)

Ra[bcd] = 0

ga[bcd] = 0,

(1.9.19)

gabcd = gcdab .

(1.9.20)

Rabcd = Rcdab

To motivate the definition, let us temporarily assume that gab is positive-definite. (That makes things a bit easier.) Let p be a point in M and let W be a two-dimensional subspace of Mp . We take the W-sectional curvature of (M, gab ) at p to be the number Rabcd αa β b αc β d (gad gbc − gac gbd ) αa β b αc β d

(1.9.21)

where αa and β a are any two vectors at p that span W . Note that the definition is well posed. First, the denominator cannot be 0, for that would violate our stipulation that αa and β a span W . (Using a more familiar notation, the point is this: if u and v are vectors such that hu, vi2 = kuk2 kvk2 , then u

and v must be linearly dependent.) Second, the expression is independent of the choice of αa and β a . For suppose that α ã and βã form a basis for W as well, with α ã = f αa + g β a and βã = h αa + k β a . Then, by (1.9.17) and (1.9.18), ˜ c β˜d Rabcd α ˜ a β˜b α (gad gbc − gac gbd ) α ã β˜b α ˜ c β˜d

= (f k − gh)2 Rabcd αa β b αc β d = (f k − gh)2 (gad gbc − gac gbd ) αa β b αc β d ,

and the factor (f k − gh)2 simply drops out. In the special case of a smooth surface in three-dimensional Euclidean space (with the metric induced on it), the sectional curvature at any point is just what we would otherwise call the “Gaussian curvature” there. (See Spivak [57], volume 2, chapter 4.) Now we show that constancy of curvature at a point can be understood to mean equality of sectional curvatures there.

80


Proposition 1.9.10. Let M be a manifold of dimension at least 2, let gab be a positive-definite metric on M , and let κ be a real number. Then Rabcd = κ (gad gbc − gac gbd ) holds at a point iff all sectional curvatures there (i.e., all W-sectional curvatures for all two-dimensional subspaces W ) are equal to κ. Proof. The “only if” half of the assertion is immediate. For the converse, assume that all sectional curvatures are equal to κ at some point p in M . Our goal is to show that the difference tensor Dabcd

Rabcd − κ (gad gbc − gac gbd )

=

vanishes at p. Note that Dabcd inherits the symmetry conditions (1.9.17) – (1.9.20). Note, as well, that (i) Dabcd αa β b αc β d = 0 for all vectors αa and β a at p. For if αa and β a are linearly independent, the claim follows from the fact that all sectional curvatures at p are equal to κ. And if they are not linearly independent, it follows from (1.9.17) (or (1.9.18)). What we show is that Dabcd cannot satisfy (i) and the listed symmetry conditions without vanishing. 1

2

n

i

j

i

k

Let µa , µa , ..., µa be a basis for Mp . We claim that (ii) Dabcd µa µb µc µd = 0, for all i, j and k. This is clear, since by (i) and (1.9.20), j

j

j

0 = Dabcd µa (µb + µb ) µc (µd + µd ) = 2 Dabcd µa µb µc µd . i

k

i

i

k

k

i

We also claim that (iii) Dacbd = −Dadbc . For this, note that by (i) and (ii), and the symmetries (1.9.17),

(1.9.18), (1.9.20), 0 = =

j

j

j

j

Dabcd (µa + µa )(µb + µb )(µc + µc )(µd + µd ) = 2 Dabcd (µa µb µc µd + µa µb µc µd ) i

k

i

l

j

k

l

i

k

l

i

l

k

2 (Dacbd + Dadbc ) µa µb µc µd . i

i

j

k

l

1

2

n

Since this holds for all µa µb µc µd (and since µa , µa , ..., µa is a basis for Mp ), we have (iii). Finally, it k

l

follows from (iii) and the other symmetries of Dabcd that Dabcd = −Dadcb = Dadbc = −Dacbd = Dadcb + Dabdc = −Dabcd − Dabcd . So Dabcd = 0. Now we drop our temporary assumption that we are dealing with a positive-definite metric and return to the general case. So far, we have considered only the property of having constant curvature at a point. We say that (M, gab ) has constant curvature if it has constant curvature at every point and the value of the curvature is everywhere the same. The second clause (same value at every point) needs to be added because it does not follow automatically — at least not if M is two-dimensional. (In that special case, the property of


81

having constant curvature at every point is vacuous and there is no reason why sectional curvatures at different points need be equal.) But, perhaps surprisingly, it does follow automatically if the dimension of M is at least 3. Proposition 1.9.11 (Schur’s Lemma). Let M be a manifold of dimension n ≥ 3, and let gab be a metric

on M (not necessarily positive-definite). Suppose there is a smooth scalar field κ on M such that Rabcd = κ (gad gbc − gac gbd ). Then κ is constant.

Proof. By Bianchi’s identity (proposition 1.8.2), ∇[m Rab cd] = 0. It follows that if we apply ∇m to

κ (δ a d δ b c − δ a c δ b d ), and anti-symmetrize over m, c, d, we get 0. But (δ a d δ b c − δ a c δ b d ) is already

anti-symmetric in c, d. So

0 = ∇[m (κ δ a d δ b c] ) = δ a [d δ b c ∇m] κ. Contracting on indices a, d and on b, c yields 0 = (n − 1) (n − 2) ∇m κ. So (given our assumption that n ≥ 3), we may conclude that ∇m κ = 0, i.e., that κ is constant on

M.

As it happens, the assertion of the proposition is also true if n = 1, for in that case we have (at every point) Rabcd = 0 = (gad gbc − gac gbd ). (Every tensor over a one-dimensional vector space vanishes if it is anti-symmetric in two indices.) The proposition fails only if n = 2.

′ Let (M, gab ) and (M ′ , gab ) be two manifolds with metric. We say they are locally isometric if, for all

points p ∈ M and p′ ∈ M ′ , there exist open sets O ⊆ M and O′ ⊆ M ′ containing p and p′ , respectively,

′ such that the restricted manifolds (O, gab|O ) and (O′ , gab|O ′ ) are isometric.

′ Suppose (M, gab ) and (M ′ , gab ) both have constant curvature and their respective curvature values

are κ and κ′ . Then, one can show, they are locally isometric iff (i) M and M ′ have the same dimension, ′ (ii) gab and gab have the same signature, and (iii) κ = κ′ . (See Wolf [64], proposition 2.4.11.) But these ′ conditions certainly do not guarantee that (M, gab ) and (M ′ , gab ) are (globally) isometric. (We will have

more to say about this in section 2.11.)


1.10

82

Hypersurfaces

Let (S, CS ) and (M, CM ) be manifolds of dimension k and n respectively, with 1 ≤ k ≤ n. A smooth

map Ψ : S → M is said to be an imbedding if it satisfies the following three conditions. (I1) Ψ is injective.

(I2) At all points p in S, the associated (push-forward) linear map (Ψp )∗ : S p → M Ψ(p) is injective. (I3) For all open sets O1 in S, Ψ[O1 ] = Ψ[S] ∩ O2 for some open set O2 in M . (Equivalently, the inverse map Ψ−1 : Ψ[S] → S is continuous with respect to the relative topology on Ψ[S].)

(Recall our discussion of push forward and pull-backward maps in section 1.5.) Several comments about the definition are in order. First, given any point p in S, (I2) implies that (Ψp )∗ [Sp ] is a k-dimensional subspace of MΨ(p) . So the condition cannot be satisfied unless k ≤ n.

Second, the three conditions are independent of one another. For example, the smooth map Ψ : R → R2

defined by Ψ(s) = (cos(s), sin(s)) satisfies (I2) and (I3) but is not injective. It wraps R round and round in a circle. On the other hand, the smooth map Ψ : R → R defined by Ψ(s) = s3 satisfies (I1) and (I3) but is not an imbedding because (Ψ0 )∗ : R0 → R0 is not injective.4 (Here R0 is the tangent space to

the manifold R at the point 0). Finally, a smooth map Ψ : S → M can satisfy (I1) and (I2) but still

have an image that “bunches up on itself”. It is precisely this possibility that is ruled out by condition (I3). Consider, for example, a map Ψ : R → R2 whose image consists of part of the image of the curve

y = sin(1/x) smoothly joined to the segment {(0, y) : y < 1}, as in figure 1.10.1. It satisfies conditions

(I1) and (I2) but is not an imbedding because we can find an open interval O1 in R such that given any open set O2 in R2 , Ψ[O1 ] 6= O2 ∩ Ψ[R]. Suppose (S, CS ) and (M, CM ) are manifolds with S ⊆ M . We say that (S, CS ) is an imbedded

submanifold of (M, CM ) if the identity map id : S → M is an imbedding. If, in addition, k = n − 1 (where k and n are the dimensions of the two manifolds), we say that (S, CS ) is a hypersurface in

(M, CM ). In what follows, we first work with arbitrary imbedded submanifolds and then restrict attention

to hypersurfaces. Where confusion does not arise, we suppress reference to charts.

Once and for all in this section, let (S, CS ) be a k-dimensional imbedded submanifold of the n-

dimensional manifold (M, CM ), and let p be a point in S. We need to distinguish two senses in which

one can speak of “tensors at p”. There are tensors over the vector space Sp (call them S-tensors at p) and ones over the vector space Mp (call them M -tensors at p). So, for example, a S-vector ξã at p ˜ → R where O ˜ is a subset of S that is open in the topology makes assignments to maps of the form f˜: O d in R0 (and so has a non-trivial kernel). This is clear since, for any smooth real-valued dx function f defined on some open subset of R containing Ψ(0) = 0, we have 4

(Ψ0 )∗ annihilates the vector

` ` d ´ ` d ´ ` ´ d ´ (Ψ0 )∗ ( ) (f ) = (f ◦ Ψ) |x=0 = (f (x3 ) |x=0 = f ′ (x3 ) 3x2 |x=0 = 0. dx dx dx

83


y Ψ ⌢ (

) O1

R

Ψ[O1 ] ⌣

x

Figure 1.10.1: The map Ψ is not an imbedding, because its image bunches up on itself.

induced by CS , and f˜ is smooth relative to CS . In contrast, an M -vector ξ a at p makes assignments

to maps of the form f : O → R where O is a subset of M that is open in the topology induced by CM ,

and f is smooth relative to CM .5 Our first task is to consider the relation between S-tensors at p and

M -tensors there.

Let us say that ξ a ∈ (Mp )a is tangent to S if ξ a ∈ (idp )∗ [(Sp )a ]. (This makes sense. We know that

(idp )∗ [(Sp )a ] is a k-dimensional subspace of (Mp )a . ξ a either belongs to that subspace or it does not.) Let us further say that ηa in (Mp )a is normal to S if ηa ξ a = 0 for all ξ a ∈ (Mp )a that are tangent to S.

Each of these classes of vectors has a natural vector space structure. The space of vectors ξ a ∈ (Mp )a

tangent to S has dimension k. The space of co-vectors ηa ∈ (Mp )a normal to S has dimension (n − k)

(see problem 1.10.1).

Problem 1.10.1. Let S be a k-dimensional imbedded submanifold of the n-dimensional manifold M , and let p be a point in S. (1) Show that the space of co-vectors ηa ∈ (Mp )a normal to S has dimension (n − k). (Hint: Consider a basis for (Mp )a containing (as a subset) k vectors tangent to S. Then consider a dual basis.)

(2) Show that a vector ξ a ∈ (Mp )a is tangent to S iff ηa ξ a = 0 for all co-vectors ηa ∈ (Mp )a that are normal to S.

We note for future reference that a co-vector ηa ∈ (Mp )a is normal to S iff (idp )∗ (ηa ) = 0. It is worth

giving the argument in detail to help gain familiarity with our notation. (idp )∗ (ηa ) is the zero vector in (Sp )a iff (idp )∗ (ηa ) ξã = 0 for all ξã ∈ (Sp )a . But (by the definition of the pull-back operation), (idp )∗ (ηa ) ξã = ηa (idp )∗ (ξã ) . So (idp )∗ (ηa ) = 0 iff ηa (idp )∗ (ξã ) = 0 for all ξã ∈ (Sp )a . But a vector ξ a ∈ (Mp )a is tangent to S precisely if it is of the form (idp )∗ (ξã ) for some ξã ∈ (Sp )a . So

(idp )∗ (ηa ) = 0 iff ηa ξ a = 0 for all vectors ξ a ∈ (Mp )a that are tangent to S, i.e., ηa is normal to S.

The classification we have introduced can be extended to indices on M -tensors of higher index structure. Consider, for example, the M -tensor αabcd at p. We take it to be tangent to S in its first contravariant 5 As

an aid to clarity, we shall sometimes mark S-tensors with a tilde, and sometimes we shall indicate the character of

a vector ξ a simply by indicating, explicitly, its membership in (Sp )a or (Mp )a . (Co-vectors ηa shall be handled similarly.)


84

index if ηa αabcd = 0 for all ηa ∈ (Mp )a that are normal to S. (Note that this characterization, which applies to all M -tensors with contravariant indices, is consistent with the one given initially for the special

case of contravariant vectors by virtue of the second assertion in problem 1.10.1.) And we take it to be normal to S in its second covariant index if ξ d αabcd = 0 for all ξ d ∈ (Mp )d that are tangent to S. So far, M -tensors at p can be tangent to S only in their contravariant indices and normal to S only in their covariant indices. But now (and henceforth in this section), let us assume that a metric gab is present on M . Then the classification can be extended. We can take take the tensor to be tangent to S in a covariant index if it is so after the index is raised with gab . And we can take it to be normal to S in a contravariant index if it is so after the index is lowered with gab . Now we have four subspaces to consider side by side. In addition to the old k-dimensional space of contravariant M -vectors at p tangent to S, we have a new (n − k)-dimensional space of contravariant M -vectors at p normal to S. And in

addition to the old (n − k)-dimensional space of covariant M -vectors at p normal to S, we have a new k-dimensional space of covariant M -vectors at p tangent to S. As one would expect, it is possible to

introduce “projection tensors” that, when applied to (contravariant and covariant) M -vectors at p, yield their respective components in these four subspaces. We shall do so in a moment. Let us say that an M -tensor at p is (fully) tangent to S (or normal to S) if it is so in each of its indices. The subspace of M -tensors αa1 ...ar b1 ...bs at p tangent to S has dimension k (r+s) . Nothing said so far rules out the possibility that there is a non-zero vector ξ a ∈ (Mp )a that is both

tangent to, and normal to, S. Such a vector would necessarily satisfy gab ξ a ξ b = 0. (Since ξ a is tangent to S, and gab ξ b is normal to S, the contraction of the two must be 0.) There cannot be non-zero vectors satisfying this condition if gab is positive definite. But the possibility does arise when, for example, the metric is of Lorentzian signature. We say that our imbedded submanifold S is a metric submanifold (relative to the background metric gab on M ) if, for all p in S, no non-zero vector in (Mp )a is both tangent to S and normal to S. An alternative formulation is available. The pull-back field id∗ (gab ) is always a smooth, symmetric field on S. But it is non-degenerate (and so a metric) iff S is a metric submanifold (see problem 1.10.2). Problem 1.10.2. Let S be a k-dimensional imbedded submanifold of the n-dimensional manifold M , and let gab be a metric on M . Show that S is a metric submanifold (relative to gab ) iff for all p in S, the pull-back tensor (idp )∗ (gab ) at p is non-degenerate, i.e., there is no non-zero vector ξã ∈ (Sp )a such that (idp )∗ (gab ) ξã = 0.

In what follows, we assume that S is a metric submanifold (relative to gab ). Non-metric submanifolds

do arise in relativity theory. (“Null hypersurfaces”, for example, are non-metric.) But they are not essential for our purposes, and it will simplify our discussion to put them aside. The assumption that S is a metric submanifold, for example, implies – and, indeed, is equivalent to the assertion that – there is a basis for Mp consisting entirely of vectors that are either tangent to, or normal to, S (but not both). It

85


is convenient to be able to work with such a basis. (It is always true (in the presence of a metric) that we can find k linearly independent vectors at p tangent to S, and (n − k) linearly independent vectors there

normal to S. But the combined set of n vectors will be linearly independent iff the subspaces spanned by the two individual sets share no non-zero vector, i.e., there is no non-zero vector that is both tangent to, and normal to, S.) The vector space of S-tensors at p of a given index structure has the same dimension as the vector space of M -tensors there that are of the same index structure and that are tangent to S. In fact, as we now show, there is a canonically defined linear map φp from the first to the second that is injective and so qualifies as an isomorphism.6 We define this isomorphism α ã1 ...ar b1 ...bs 7−→ φp (˜ αa1 ...ar b1 ...bs )

in stages, considering, in order, scalars (0th order tensors), contravariant vectors, covariant vectors, and

then, finally, arbitrary tensors. For scalars α, we set φp (α) = α. (We do not put a tilde over the first α because there is no distinction to be drawn here. Scalars are just scalars.) For vectors ξã , we set φp (ξã ) = (idp )∗ (ξã ). It follows immediately from (I2) — the second condition in the definition of an imbedding — and the definition of tangency that φp determines an isomorphism between Sp and the space of contravariant M -vectors at p tangent to S. Next, we define φp (˜ ηa ) by specifying its action on vectors ξ a ∈ (Mp )a that

are either tangent to, or normal to, S. (This suffices since, as we have seen, we can always find a basis for (Mp )a consisting entirely of such vectors.)   ηã (φp )−1 (ξ a ) a φp (˜ ηa ) ξ =  0

if ξ a is tangent to S if ξ a is normal to S

Clearly, φp (˜ ηa ) is tangent to S. That much is guaranteed by the second clause within the definition. Moreover, the action of φp on (Sp )a is injective. (Suppose φp (˜ ηa ) is the zero vector in (Mp )a . Then −1 a a a ηã (φp ) (ξ ) = 0, for all tangent vectors ξ ∈ (Mp ) . But every vector in (Sp )a is of the form

(φp )−1 (ξ a ) for some tangent vector ξ a ∈ (Mp )a . So ηã is the zero vector in (Sp )a .)

Finally, we consider the case of an S-tensor at p of higher order index structure – say α ãbc . There are no surprises. We define φp (˜ αabc ), once again, by specifying its action on vectors that are all tangent to, or normal to, S. ab

φp (˜ α

c ) µa

c

νb ξ =

  α ˜ ab  0

c

(φp )−1 (µa ) (φp )−1 (νb ) (φp )−1 (ξ c ) if µa , νb , and ξ c are tangent to S if µa , νb , or ξ c is normal to S

Clearly, φp (˜ αabc ) is tangent to S, and the argument that φp is injective in its action on (Sp )ab c is very much the same as in the preceding case. This completes our definition. 6 We

are presenting a great deal of detail here. Some readers may want to skip to proposition 1.10.1.

86


We have established, so far, that for every index structure a1 ...ar b1 ...bs , there is an isomorphism between the vector space of S-tensors α ˜ a1 ...ar b1 ...bs at p and the vector space of M -tensors αa1 ...ar b1 ...bs at p that are tangent to S. If we now “aggregate” the different isomorphisms, we arrive at a map φp (we use the same notation) that commutes with all the tensor operations – addition, exterior multiplication, index substitution, and contraction. It follows from our definition, for example, that φp (˜ αab β˜de ) = c

αabc ) φp (β˜ce ). In summary, we have established the following. φp (˜ α c ) φp (β˜de ) and φp (˜ αabc β˜ce ) = φp (˜ ab

Proposition 1.10.1. Let S be a metric submanifold of the manifold M . Then the tensor algebra of S-tensors at any point of S is isomorphic to the tensor algebra of M -tensors there that are tangent to S. The map φp is closely related to (idp )∗ . Indeed, it agrees with the latter in its action on contravariant tensors at p. But (idp )∗ makes assignments only to contravariants tensors there, whereas φp makes assignments to all tensors. (Similarly, (φp )−1 agrees with (idp )∗ in its assignment to covariant tensors at p that are tangent to S.) Now we switch our attention to tensor fields on S, i.e., assignments of tensors of the same index structure to every point of S. Of course, we have to distinguish between assignments of S-tensors and assignments of M -tensors. But the isomorphisms we have been considering (defined at individual points of S) induce a correspondence α ˜ a1 ...ar b1 ...bs 7−→ φ(˜ αa1 ...ar b1 ...bs ) between S-fields and M -fields that are tangent to S, i.e., tangent at every point.

The correspondence respects differential structure in the following sense (in addition to algebraic structure). Let αa1 ...ar b1 ...bs be an M -field on S that is tangent to S. There are two senses in which it might be said to be “smooth”. Let us say that it is M -smooth if, for every p in S, there is an open set +

O ⊆ M containing p and an extension of αa1 ...ar b1 ...bs to a field αa1 ...ar b1 ...bs on O that is smooth relative

to the charts CM . (This sense of smoothness applies to all M -fields on S, whether they are tangent to

S or not.) Let us also say that it is S-smooth if the corresponding S-field φ−1 (αa1 ...ar b1 ...bs ) is smooth relative to the charts CS . One would like these two senses of smoothness to agree, and in fact they do.

By direct consideration of charts, one can establish the following. (We skip the proof.)

Proposition 1.10.2. Let S be a metric submanifold of the manifold M . Further, let αa1 ...ar b1 ...bs be an M -field on S that is tangent to S. Then αa1 ...ar b1 ...bs is M -smooth iff it is S-smooth. In what follows, we shall sometimes say that an M -field on S is smooth without further qualification. If the field is not tangent to S, this can only mean that it is M -smooth. If it is tangent to S, the proposition rules out any possibility of ambiguity. ˜ ab = id∗ (gab ) on S. It is called the induced metric or first fundamental form Consider now the S-field h on the manifold S.7 (That it is a metric follows from our assumption that S is a metric submanifold of 7 Warning:

˜ ab is not a “form” in the special the latter (perfectly standard) expression is potentially confusing because h

technical sense introduced in section 1.7, i.e., is it not anti-symmetric.

87


˜ ab is a unique compatible derivative operator D ˜ on S. (So M . Recall problem 1.10.2.) Associated with h ˜ bc = 0.) It is our goal now to show that it is possible, in a sense, to express D ˜ in terms of ã h it satisfies D the derivative operator ∇ on M that is compatible with gab . The sense involved is a bit delicate because

it involves the map φ we have been considering that takes S-fields to M -fields on S tangent to S. The idea, in effect, is to translate talk about the former into talk about the latter. ˜ ab is a smooth, symmetric M -field hab = φ(h ˜ ab ) = φ(id∗ (gab )) on S that is tangent Corresponding to h to S. (It is tangent to S because the image of every S-field under φ is so. How do we know it is smooth? Since gab is a smooth field on the manifold M , id∗ (gab ) is a smooth field on the manifold S. But id∗ (gab ) = φ−1 (hab ). So hab is S-smooth (and, hence, M -smooth as well).) We can characterize hab ˜ ab or φ, in terms of its action (at any point of S) on M -vectors that are directly, without reference to h tangent to, or normal to, S. hab λa η b =

   gab λa η b  0

if λa and η a are both tangent to S

(1.10.1)

if λa or η a is normal to S.

The equivalence is easy to check.8

Several properties of hab , as well as a companion field kab = (gab − hab ) are listed in the following

proposition. Clearly, kab is also a symmetric, smooth M -field on S. (Here and in what follows, whenever we lower and raise indices on M -tensors, it should be understood that we do so with gab .) Proposition 1.10.3. Let S be a metric submanifold of the manifold M (with respect to the metric gab on M ). Let hab be the M -field on S defined by (1.10.1), and let kab be the companion M -field (gab − hab )

on S. Then all the following hold.

(1) hab is tangent to S and kab is normal to S. (2) For all M -vector fields αa on S, (a) αa is tangent to S ⇐⇒ hab αb = αa ⇐⇒ k ab αb = 0; (b) αa is normal to S ⇐⇒ k ab αb = αa ⇐⇒ hab αb = 0. (3) hab hbc = hac and k ab k bc = k ac and hab k bc = 0. Proof. (1) We have already given an argument to show that hab is tangent to S. (Once again, hab = ˜ ab ), and the image of every S-field under φ is tangent to S.) Now let ξ a be any M -vector tangent φ(h to S (at any point of S). Then kab ξ a = gab ξ a − hab ξ a . But gab ξ a = hab ξ a , since they agree in their

action on both vectors tangent to S and normal to it. So kab ξ a = 0. It follows that kab is normal to S

in its first index. But kab is symmetric. So it is (fully) normal to S. (2) Suppose first that hab αb = αa . 8 Suppose

λa and ηa are both tangent to S. Then, by the definitions of φ and the pull-back map id∗ ,

hab λa ηb = φ(id∗ (gab )) λa ηb = id∗ (gab ) φ−1 (λa ) φ−1 (ηb ) = gab id∗ (φ−1 (λa )) id∗ (φ−1 (ηb )) = gab λa ηb . Alternatively, if either λa or ηa is normal to S, then hab λa ηb = 0 since hab is tangent to S.

88


Then αa is certainly tangent to S, since hab is tangent to S in the index a. Conversely, suppose αa is tangent to S. Then, we claim, hab αb and αa have the same action on any vector ηa (at any point of S) that is either tangent to, or normal to, S. In the first case, hab αb ηa = gab η a αb = αa ηa . In the second case, hab αb ηa = 0 = αa ηa . This gives us the first equivalence in (a). The second is immediate since k ab αb = (g ab − hab ) αb = αa − hab αb . The equivalences in (b) are handled similarly. (3) It follows from

(2) that hab hbc and hac have the same action on any vector ξ c (at any point of S) that is either tangent to, or normal to, S. So hab hbc = hac . The arguments for k ab k bc = k ac and hab k bc = 0 are similar. Problem 1.10.3. Prove the following generalization of clause (2) in proposition 1.10.3. For all M -tensor fields α ...a... on S, (1) α ...a... is tangent to S in the index a ⇐⇒ hab α ...b... = α ...a... ⇐⇒ k ab α ...b... = 0. (2) α ...a... is normal to S in the index a

⇐⇒ k ab α ...b... = α ...a... ⇐⇒ hab α ...b... = 0.

We have formulated the preceding problem in terms of contravariant M -fields on S. But, of course, this involves no essential loss of generality. For given one, instead, of form, say, α ab cde , we can always apply the stated results to β abcde = α ab mnr g mc g nd g re and then lower indices. We can think of hab and k ab as projection operators. Given an M -field ξ a , hab ξ b is its component tangent to S, and k ab ξ b is its component normal to S. More generally, we can use the two operators to decompose an M -field of arbitrary index structure into a sum of component tensor fields, each of which is either tangent to S or normal to S in each of its indices (which is not to say that each of the component fields will be either (fully) tangent to S or (fully) normal to S). So, for example, in the case of a field αab on S, we have the following decomposition: αab = ham hnb αmn + ham k nb αmn + k am hnb αmn + k am k nb αmn . (Notice that the two fields, left and right, have the same action (at any point) on any pair of vectors ηa ξ b , each of which is either tangent to S or normal to S.) ˜ can be expressed in terms of ∇. We start We are ready to explain the sense in which the action of D

with a lemma.

Lemma 1.10.4. Let S be a metric submanifold of the manifold M (with respect to the metric gab on 1

2

M ). Let hab be the M -field on S defined by (1.10.1). Finally, let α a1 ...ar b1 ...bs and α a1 ...ar b1 ...bs be smooth M -fields on an open set O ⊆ M that agree on S. Then at all points of S ∩ O, 2

1

hnm ∇n α a1 ...ar b1 ...bs = hnm ∇n α a1 ...ar b1 ...bs . 1 a1 ...ar

Proof. Consider β a1 ...ar b1 ...bs = α S ∩ O. We need to show that

b1 ...bs −

2 a1 ...ar

α

b1 ...bs .

hnm ∇n β a1 ...ar b1 ...bs = 0

It vanishes on S. Let p be a point in

89


at p. To do so, it suffices to show that if we contract the left side with any vector ξ m at p that is either tangent to, or normal to, S, the result is 0. That is true automatically if ξ m is normal to S (since hnm is tangent to S). And if ξ m is tangent to S, hnm ξ m = ξ n . So it suffices to show that ξ n ∇n β a1 ...ar b1 ...bs = 0 for all ξ n at p tangent to S. The proof of this assertion is similar to other “well-definedness” arguments given before, and proceeds by considering the index structure of β a1 ...ar b1 ...bs .

If β is a scalar field on

S, then ξ n ∇n β is just the directional derivative ξ(β). This has to be 0 because β is constant on S.

One next proves the statement for contravariant vector fields β a on S using the result for scalar fields together with the Leibniz rule, and so forth. Now suppose αa1 ...ar b1 ...bs is a smooth M -field on S. We cannot expect to be able to associate with it a field ∇m αa1 ...ar b1 ...bs on S. (The latter, if well defined, would encode information about how αa1 ...ar b1 ...bs

changes as one moves away from S in arbitrary directions.) But, by the lemma, we can introduce a field 1

hnm ∇n αa1 ...ar b1 ...bs on S. At any point of S, we simply extend αa1 ...ar b1 ...bs to a smooth field α a1 ...ar b1 ...bs

on some open set O, and set

1

hnm ∇n αa1 ...ar b1 ...bs = hnm ∇n α a1 ...ar b1 ...bs . This field need not be tangent to S even if αa1 ...ar b1 ...bs is. But we can “make it tangent” if we project all indices onto S with the field hnm . This action defines an operator Da on the set of all smooth M -tensor fields αa1 ...ar b1 ...bs on S that are tangent to S: Dm αa1 ...ar b1 ...bs = ha1c1 ... harcs hd1b1 ... hdsbs hnm ∇n αc1 ...cr

d1 ...ds .

(1.10.2)

The basic result toward which we have been working is the following. Proposition 1.10.5. For all smooth S-fields α ˜ a1 ...ar b1 ...bs , ñ α φ(D ã1 ...ar b1 ...bs ) = Dn φ(˜ αa1 ...ar b1 ...bs ). ˜ be the operator on smooth S-fields that is defined by the condition ˜ Proof. Let D ˜ ñ α D ã1 ...ar b1 ...bs = φ−1 (Dn φ(˜ αa1 ...ar b1 ...bs )). ˜ ab . For then It suffices for us to show that it is a derivative operator on S and that is compatible with h ˜˜ = D. ˜ it will follow (by proposition 1.9.2) that D ˜ ab ) = hab , we have Consider, first, the compatibility condition. Since φ(h ˜ ˜ ab ñ h D

˜ ab )) = φ−1 (Dn hab ) = φ−1 (hr hs hm ∇m hrs ) = φ−1 (Dn φ(h a b n r r −1 s m = φ (h b h n ∇m (h a hrs ) − hrs ∇m h a )

= φ−1 (hsb hmn ∇m has − hrb hmn ∇m hra ) = φ−1 (0) = 0.

90


Note that we have used the Leibniz rule (in reverse) to arrive at the fourth equality. We are justified in doing so because we are here working “within the shadow” of the projection operator hmn . We can always (locally) extend the tensor fields in question, invoke the Leibniz rule for ∇ in its standard form

(where we are working with fields defined on open sets in M rather than fields defined only on S), and then invoke our lemma to show that it does not matter how we do the extension. Note also that the fifth equality follows from the third clause of proposition 1.10.3, and the sixth from the symmetry of hab .

˜ ˜ satisfies conditions (DO1) through (DO6). The first five are straightforNext we need to verify that D ward. The argument is very much the same in each case. Let us consider, for example, a representative instance of the Leibniz rule. We have ˜ ˜ n (˜ D αab η˜c ) = = =

φ−1 (Dn [φ(˜ αab ) φ(˜ ηc )]) = φ−1 (har hbs hqc hmn ∇m [φ(˜ αrs ) φ(˜ ηq )])

φ−1 (har hbs hqc [φ(˜ αrs ) hmn ∇m φ(˜ ηq ) + φ(˜ ηq ) hmn ∇m φ(˜ αrs )])

αrs )) ηq ) + φ(˜ ηc ) har hbs hmn ∇m φ(˜ φ−1 (φ(˜ αab ) hqc hmn ∇m φ(˜

=

φ−1 (φ(˜ αab ) Dn φ(˜ ηc ) + φ(˜ ηc ) Dn φ(˜ αab ))

=

α ˜ ab φ−1 (Dn φ(˜ ηc )) + η˜c φ−1 (Dn φ(˜ αab )) ˜˜ α ˜ η˜ + η˜ D ˜ ˜ ab . α ˜ ab D

=

n c

c

n

A few steps here deserve comment. For the fourth equality, we need the fact that har φ(˜ αrs ) = φ(˜ αas ) (and a number of similar statements involving change of index). Note that this is just an instance of the assertion in problem 1.10.3, since φ(˜ αrs ) is tangent to S. And the sixth equality holds because φ (acting at any point in S) is a tensor algebra isomorphism that commutes with the operations of addition and exterior multiplication. Let us turn, finally, to (DO6). This is the only one of the conditions that requires a bit of attention. Let α be a smooth scalar field on S. Then ˜ ˜ ˜b α = ã D D

˜ ˜ b α)) = φ−1 (Da Db φ(α)) = φ−1 (hm hn ∇n (hr ∇r α)). φ−1 (Da φ(D b a m

Here we have used the fact that φ(α) = α. Now let p be any point on S. We can extend α to a smooth +

+

field α on an open set O in M containing p. Moreover, we can do so in such a way that ∇a α is tangent +

to S on S ∩ O. (This can be verified with an argument involving charts. Intuitively we keep α constant as +

+

+

we move out from S in directions normal to S). So hna ∇n α = ∇a α on S ∩ O. Thus, ∇a α is a smooth +

field on O that agrees with hna ∇n α on S ∩ O. It follows that we can understand hna ∇n (hr m ∇r α) to +

be hna ∇n ∇m α on S ∩ O, and therefore

+ ˜ ˜˜ α = φ−1 (hm hn ∇ ∇ α ã D ) D b b a n m

at p. The tensor on the right side is manifestly symmetric in a and b (since ∇ satisfies condition (DO6)). ˜˜ D ˜ ˜ Thus D a b α is symmetric in these indices at our arbitrary point p in S.


91

Up to this point we have been attentive to the distinction between S-fields and M -fields on S tangent ˜ and D. But it is, more or less, standard practice to S, between ˜ hab and hab , and between the operators D to be a bit casual about these distinctions or even to collapse them entirely by identifying the vector space Sp formally with the subspace of Mp whose elements are tangent to S. (The work we have done to this point — in particular, propositions 1.10.1, 1.10.2, and 1.10.5 — makes clear that there is no harm in doing so.) In what follows, that will be our practice as well. We shall refer to hab as the “metric induced on S” (or the “first fundamental form on S”), refer to D as the “derivative operator induced on S”, and so forth. We shall also drop the labels “S-field” and “M -field”, since it is only the latter with which we shall be working. In effect, we shall be systematically translating “S-talk” into “M -talk”. Here is one more example of how this works. What should we mean by a “geodesic on S with respect to the induced metric (or induced derivative operator)”? We can certainly understand it to be a map of the form γ : I → S that ˜ n ξã = 0. Instead, we shall drop is smooth with respect to CS and whose tangent field ξã satisfies ξñ D ˜ and take it to be a map of the form γ : I → S that is smooth with respect explicit reference to CS and D to CM and whose tangent field ξ a satisfies ξ n Dn ξ a = 0. We know that hma hnb hp c ∇m hnp = 0

(1.10.3)

˜ bc = 0. ˜ ah on S. (This is just the assertion that Da hbc = 0, and we proved it in the course of showing that D That was the first step in our proof of proposition 1.10.5.) Similarly one can show that hma k nb k p c ∇m hnp = 0

(1.10.4)

on S. However, the mixed projection field πabc defined by πabc = hma hnb k pc ∇m hnp

(1.10.5)

need not vanish. It turns out that πabc is of particular geometric interest. It is called the extrinsic curvature field on S. Problem 1.10.4. Prove (1.10.4). The induced metric hab and its associated derivative operator D are geometric structures “intrinsic” to S. They are not sensitive to the way S is imbedded in M . We say that (S, hab ) has vanishing intrinsic curvature just in case D is flat. The extrinsic curvature of S, in contrast, is determined by the imbedding. Think of both a plane and a cylinder imbedded in ordinary three-dimensional Euclidean space (figure 1.10.2). They both have vanishing intrinsic curvature. But only the plane has vanishing extrinsic curvature. Notice that all geodesics of the plane are necessarily geodesics of the ambient threedimensional space. But the corresponding statement for the cylinder is not true. There are geodesics of the cylinder (e.g., γ in figure 1.10.2) that are not geodesics of the larger space. This is a good way to

92


γ

Figure 1.10.2: The cylinder and the plane (imbedded in three-dimensional Euclidean space) both have vanishing intrinsic curvature. But the cylinder, in contrast to the plane, has nonvanishing extrinsic curvature. Notice that there are curves on the cylinder, e.g., γ, that are geodesics with respect to induced derivative operator D that are not geodesics with respect to the background derivative operator ∇. think about extrinsic curvature. Indeed, as we shall prove (proposition 1.10.7), πabc is a measure of the degree to which geodesics in (S, hab ) fail to be geodesics in (M, gab ). But first we need a lemma. Lemma 1.10.6. π[ab]c = 0 . Proof. Consider any point p in S. If ξ a is a vector at p tangent to S, we have ξ c k pc = 0 and hence ξ c π[ab]c = 0. So it will suffice to show ξ c π[ab]c = 0 for all ξ a at p normal to S. Since S has dimension k and M has dimension n, we can find an open set O containing p and (n − k) smooth scalar fields i

i

i

α (i = 1, ..., n − k) on O such that (i) ∇a α is normal to S on S ∩ O, for all i, and (ii) the vectors ∇a α

are linearly independent on S ∩ O. (This can be verified with an argument involving charts. Indeed, i

the fields α can be local coordinates induced by a chart on M . What is required is that their associated

coordinate curves all be orthogonal to S where they intersect it.) To complete the proof, it suffices to i

verify that (∇c α) π[ab]c = 0 at p for all i. But this follows since we have: (∇c α) π[ab]c i

= (∇c α) hm[a hnb] k pc ∇m hnp = hm[a hnb] (∇p α)∇m hnp i i = hm[a hnb] [∇m (hnp ∇p α) − hnp ∇m ∇p α i

i

= −hm[a hpb] ∇m ∇p α = −hma hpb ∇[m ∇p] α = 0. i

i

i

(For the fourth equality we have used the fact that since hnp ∇p α = 0 on S ∩ O, it must be the case i

that ham ∇m (hnp ∇p α) = 0 on S ∩ O. This follows, once again, from lemma 1.10.4.) Now we can give the promised geometric interpretation of πabc .

Proposition 1.10.7. Let S be a metric submanifold of the manifold M (with respect to the metric gab on M ). Let ∇ be the derivative operator on M determined by gab , let hab be the induced metric on S,

93


and let πabc be the extrinsic curvature field on S. Finally, let γ be a geodesic in (S, hab ) with tangent field ξ a . Then ξ n ∇n ξ c = πabc ξ a ξ b .

(1.10.6)

Proof. By hypothesis, ξ n Dn ξ c = 0. And ξ n hrn = ξ r , since ξ a is tangent to S. So 0 = ξ n hrn hcm ∇r ξ m = ξ r (g cm − k cm )∇r ξ m = ξ r ∇r ξ c − k cm ξ r ∇r ξ m . Therefore, ξ r ∇r ξ c

=

k cm ξ r ∇r ξ m = k cm ξ r ∇r (hmp ξ p )

=

k cm ξ r (hmp ∇r ξ p + ξ p ∇r hmp ) = k cm ξ r ξ p ∇r hm p

=

k cm (ξ a hr a )(ξ b hp b )∇r hmp = ξ a ξ b hr a hp b k cm ∇r hpm = ξ a ξ b πabc .

Here we use the fact that k c m hm p = 0 for the fourth equality. Given any point p in S, πabc vanishes there iff πabc ξ a ξ b = 0 for all vectors ξ a at p that are tangent to S. (This follows, since πabc is symmetric in its first two indices (lemma 1.10.6) and also tangent to S in them. Recall proposition 1.4.3.) But given any vector ξ a at p tangent to S, there is a geodesic in (S, hab ) that passes through p, whose tangent vector there is ξ a . So it follows from our proposition that πabc = 0 iff all geodesics in (S, hab ) are geodesics in (M, gab ). Moreover, the requirement that equation (1.10.6) hold for all geodesics in (S, hab ) uniquely determines πabc . Next we consider the Gauss-Codazzi equations. Proposition 1.10.8. Suppose (M, gab ) and (S, hab ) are as in proposition 1.10.7, and D is the derivative operator on S determined by hab . Further suppose Ra bcd is the Riemann curvature field on M associated with ∇, and Rabcd is the Riemann curvature field on S associated with D. Then Ra bcd hm[a hnb] hpc k rd ∇m πnpr

= −2 π a[cm πd]bm + ham hnb hpc hrd Rm npr , 1 m n p r = h h h k Rmnpr . 2 a b c d

(1.10.7) (1.10.8)

Proof. The argument consists of a long computation. First, let λa be any smooth vector field on S tangent to S. Then Rabcd must satisfy 1 − Rabcd λb 2

= =

D[c Dd] λa = hp[c hrd] has ∇p (hmr hsn ∇m λn ) hp[c hrd] has (∇p hmr ) hsn ∇m λn

Now, by (1.10.3) and (1.10.5),

+ hmr (∇p hsn ) ∇m λn + hmr hsn ∇p ∇m λn .

hpc hrd ∇p hmr = hpc hrd g mq ∇p hqr = hpc hrd (k mq + hmq )∇p hqr = πcdm .

(1.10.9)

(1.10.10)


94

So, by lemma 1.10.6, the first term on the right side of (1.10.9) vanishes. The second and third terms can be simplified by using (1.10.10), the symmetry of hsn , and the fact that hr d hm r = hm d . We have 1 − Rabcd λb 2

=

hp[c hmd] has (∇p hsn ) ∇m λn + hp[c hmd] han ∇p ∇m λn

=

π[c an hmd] ∇m λn + hpc hmd han ∇[p ∇m] λn 1 π[c an hmd] ∇m λn − hpc hmd han Rn bpm λb . 2

=

Now πcan is normal to S in its third index. So πcan hnb = 0 and, therefore, πcan hmd ∇m λn

So we have, all together,

=

πcan hmd ∇m (hnb λb )

=

πcan hmd λb ∇m hnb = πcan hmd hbr λr ∇m hnb

=

πcan πdrn λr = π ac n πdrn λr .

1 1 − Rarcd − π a[cn πd]rn + han hpc hmd Rnrpm λr = 0. 2 2

Now let η b be an arbitrary smooth field on S and take hrb η b for λr . Then, since the first two terms are tangent to S in the index r, we have (Rabcd + 2 π a[cn πd]bn − han hrb hpc hmd Rnrpm )η b = 0. Since this holds for all smooth fields η b on S, the field in parentheses must vanish. This gives us (1.10.7). The second computation is similar, and we leave it as an exercise. Problem 1.10.5. Derive the second Gauss-Codazzi equation (1.10.8). The first Gauss-Codazzi equation expresses the intrinsic Riemann curvature tensor field Rabcd in terms

of the extrinsic curvature field πabc and the full background Riemann curvature field Rabcd . We shall return to it later when we consider the geometric significance of Einstein’s equation. So far we have assumed only that S is a metric submanifold of M . Let us now consider the special case where S is a metric hypersurface, i.e., has dimension k = (n − 1). A slight simplification results.

The vector space of vectors normal to S is now one-dimensional at every point of S. So it consists of multiples of some (normalized) vector ξ a where ξ a ξa = ±1. (ξ a ξa cannot be 0 precisely because S is a metric submanifold.) Whether the value of ξ a ξa is +1 or −1 depends on S and the signature of gab . At

least if S is connnected, the value will be the same at every point of S, i.e., everywhere +1 or everywhere −1. Let us assume that S and gab are such that the value is +1 at all points of S. (The other case is handled similarly.) So there are exactly two vectors ξ a normal to S at every point satisfying ξ a ξa = 1. Locally, at least, we can always make a choice so as to generate a smooth field. We say that S is two sided if it is possible to do so globally.

95


Let ξ a be one such (local or global) smooth normal field on S satisfying ξ a ξa = 1. Then hab

=

(gab − ξa ξb ),

(1.10.11)

kab

=

ξa ξb .

(1.10.12)

(Note that hab and (gab − ξa ξb ) have the same action on ξ a and on all vectors tangent to S.) Now consider the field πab defined by

πab = −πabc ξ c .

(1.10.13)

When hypersurfaces are under discussion, it (rather than πabc ) is often called the extrinsic curvature field (relative to ξ a ). It is also called the second fundamental form on S (relative to ξ a ). Notice that π[ab]

=

0,

(1.10.14)

πabc

=

−πab ξc ,

(1.10.15)

πab

=

hma hnb ∇m ξn .

(1.10.16)

The first assertion follows immediately from lemma 1.10.6. For the second, it suffices to observe that πabc and −πab ξc agree in their action on ξ c and on all vectors η c tangent to S. For the third, we have πab

= =

−πabc ξ c = −hma hnb k pc ξ c ∇m hnp = −hma hnb ξ p ∇m hnp −hma hnb [∇m (ξ p hnp ) − hnp ∇m ξ p ] = hma hpb ∇m ξp .

Equation (1.10.16) leads to an alternative interpretation of extrinsic curvature in the case of hyper+

surfaces. Let ξ a be an extension of ξ a to a smooth field of unit length on some open set O in M , and let +

+ +

+

+

+

+

hab be defined by hab = gab − ξ a ξ b . (So hab is an extension of hab and hab ξ a = 0.) Then we have: πab =

+ 1 £+ hab 2 ξ

(1.10.17)

+

on S ∩ O. To prove this, observe first that on S ∩ O £+ hab is tangent to S (in both indices). This ξ

+ a

follows since £+ ξ = 0 and, hence, ξ

+ a

+

+

+

+

+

ξ £+ hab = £+ (hab ξ a ) − hab £+ ξ a = 0. ξ

ξ

ξ

Therefore on S we have +

£+ hab ξ

=

+ +

+

hra hsb £+ hrs = hra hsb £+ (grs − ξ r ξ s ) ξ

ξ

+ n

+

+

=

hra hsb £+ grs = hra hsb [ ξ ∇n grs + gns ∇r ξ n + grn ∇s ξ n ]

=

hra

=

2 π(ab) = 2 πab .

ξ

hsb

+

+

+

(∇r ξ s + ∇s ξ r ) = 2 hr(a hsb) ∇r ξ s


96

(The final two equalities follow, respectively, from equations (1.10.16) and (1.10.14).) 1 Thus we can think of πab (up to the factor ) as the Lie derivative of hab in the direction ξ a normal 2 to S. This interpretation will be important later in connection with our discussion of the “initial value problem” in general relativity. Finally let us reconsider the Gauss-Codazzi equations in the present case. Substituting −πab ξc for

πabc in the equations of proposition 1.10.8 yields Rabcd

=

hm[a hnb] hpc ∇m πnp

=

−2 π a[c πd]b + ham hnb hpc hrd Rm npr , 1 − hma hnb hpc ξ r Rmnpr . 2

(1.10.18)

The second can be expressed as 1 D[a πb]c = − hma hnb hpc ξ r Rmnpr . 2

(1.10.19)

Contracting on (1.10.18) yields Rbc = −π ac πab + π aa πcb + hrm hnb hpc Rm npr . Substituting (g rm − ξm ξ r ) for hrm on the right side, we arrive at Rbc = π πbc − πab π ac + hnb hpc Rnp − Rmbcr ξ m ξ r ,

(1.10.20)

where π = π aa . (Note that Rmbcr ξ m ξ r is tangent to S, since (by proposition 1.9.4) contracting with ξ a on b or c yields 0. Hence hnb hpc Rmbcr ξ m ξ r = Rmbcr ξ m ξ r .) Contracting once more yields R

= π 2 − πab π ab + hnp Rnp − Rmr ξ m ξ r = π 2 − πab π ab + R − 2 Rnr ξ n ξ r .

(1.10.21)

Of course, in the special case where we are dealing with a hypersurface imbedded in a flat manifold (Rabcd

= 0), e.g., in the case of a two-dimensional surface imbedded in three-dimensional Euclidean space,

our expressions for Rabcd , Rbc , and R simplify still further: Rabcd Rbc R

= πad πbc − πac πbd ,

(1.10.22)

= π πbc − πab π ac ,

(1.10.23)

= π 2 − πab π ab .

(1.10.24)

97


1.11

Volume Elements

In what follows, let M be an n−dimensional manifold (n ≥ 1). As we know from section 1.7, an

s-form on M (s ≥ 1) is a covariant field αb1 ...bs that is anti-symmetric (i.e., anti-symmetric in each pair of indices). The case where s = n is of special interest. i

Let αb1 ...bn be an n-form on M . Further, let ξb (i = 1, ..., n) be a basis for the tangent space at a point i

in M with dual basis η b (i = 1, ..., n). Then αb1 ...bn can be expressed there in the form 1

n

αb1 ...bn = k n! η [b1 ... η bn ]

(1.11.1)

where 1

n

k = αb1 ...bn ξ b1 ... ξ bn . (To see this, observe that the two sides of (1.11.1) have the same action on any collection of n vectors 1

n

from the set {ξb , ..., ξb }.) It follows that if αb1 ...bn and βb1 ...bn are any two smooth, non-vanishing n-forms

on M , then

βb1 ...bn = f αb1 ...bn for some smooth non-vanishing scalar field f . Smooth, non-vanishing n-forms always exist locally on M . (Suppose (U, ϕ) is a chart with coordinate 1

i

n

vector fields (~γ1 )a , ..., (~γn )a , and suppose η b (i = 1, ..., n) are dual fields. Then η [b1 ...η bn ] qualifies as a smooth, non-vanishing n-form on U .) But they do not necessarily exist globally. Suppose, for example, that M is the two-dimensional M¨ obius strip (see figure 1.11.1), and αab is any smooth two-form on M . We see that αab must vanish somewhere as follows. Let p be any point on M at which αab 6= 0, and let ξ a be any non-zero vector at p. Consider the

number αab ξ a ρb as ρb rotates though the vectors in Mp . If ρb = ±ξ b , the number is zero. If ρb 6= ±ξ b ,

the number is non-zero. Therefore, as ρb rotates between ξ a and −ξ a , it is always positive or always negative. Thus αab determines a “positive direction of rotation” away from ξ a on Mp . αab must vanish

somewhere because one cannot continuously choose positive rotation directions over the entire M¨ obius strip. M is said to be orientable if it admits a (globally defined) smooth, non-vanishing n-form. So far we have made no mention of metric structure. Suppose now that our manifold M is endowed with a metric gab of signature (n+ , n− ). We take a volume element on M (with respect to gab ) to be a smooth n-form ǫb1 ...bn that satisfies the normalization condition −

ǫb1 ...bn ǫb1 ...bn = (−1)n n!.

(1.11.2)

i

Suppose ǫb1 ...bn is a volume element on M , and ξb (i = 1, ..., n) is an orthonormal basis for the tangent space at a point in M . Then at that point we have, by (1.11.1), 1

n

ǫb1 ...bn = k n! ξ [b1 ...ξ bn ]

(1.11.3)

98


Mp

ρa

•

ξ

αab ξ a ρb > 0

a

αab ξ a (−ρb ) < 0

p

Figure 1.11.1: A 2-form αab on the M¨ obius strip determines a “positive direction of rotation” at every point where it is non-zero. So there cannot be a smooth, non-vanishing 2-form on the M¨ obius strip.

1

n

where k = ǫb1 ...bn ξ b1 ...ξ bn . Hence, by the normalization condition (1.11.2), −

n 1 n 1 k n! ξ [b1 ...ξ bn ] k n! ξ [b1 ...ξ bn ] − 1 1 1 b1 n n bn = k 2 (n!)2 ξ b1 ξ ... ξ bn ξ = k 2 n! (−1)n . n!

(−1)n n! =

So k 2 = 1 and, therefore, (1.11.3) yields

1

n

ǫb1 ...bn ξ b1 ... ξ bn = ±1.

(1.11.4)

Clearly, if ǫb1 ...bn is a volume element on M , then so is −ǫb1 ...bn . It follows from the normalization condition (1.11.4) that there cannot be any others. Thus, there are only two possibilities. Either (M, gab ) admits no volume elements (at all) or it admits exactly two, and these agree up to sign. Condition (1.11.4) also suggests where the term “volume element” comes from. Given arbitrary 1

n

1

n

vectors γ a , ..., γ a at a point, we can think of ǫb1 ...bn γb1 ...γbn as the volume of the (possibly degenerate) parallelepiped determined by the vectors. Notice that, up to sign, ǫb1 ...bn is characterized by three properties. (VE1) It is linear in each index. (VE2) It is anti-symmetric. (VE3) It assigns a volume V with |V | = 1 to each orthonormal parallelepiped. These are conditions we would demand of any would-be volume measure (with respect to gab ). If the length of one edge of a parallelepiped is multiplied by a factor k, then its volume should increase by that factor. And if a parallelepiped is sliced into two parts, with the slice parallel to one face, then its volume should be equal to the sum of the volumes of the parts. This leads to (VE1). Furthermore, if any two edges of the parallelepiped are coalligned (i.e., if it is a degenerate parallelepiped), then its volume should be zero. This leads to (VE2). (If for all vectors ξ a , ǫb1 ...bn ξ b1 ξ b2 = 0, then it must be the case

99


that ǫb1 ...bn is anti-symmetric in indices (b1 , b2 ). And similarly for all other pairs of indices.) Finally, if the edges of a parallelepiped are orthogonal, then its volume should be equal to the product of the lengths of the edges. This leads to (VE3). The only unusual thing about ǫb1 ...bn as a volume measure is 1

n

that it respects orientation. If it assigns V to the ordered sequence γ a , ..., γ a , then it assigns (−V ) to 2

1

3

n

γ a , γ a , γ a , ..., γ a , and so forth. It will be helpful to collect here a few facts for subsequent calculations. Suppose ǫa1 ...an is a volume element on M with respect to the metric gab with signature (n+ , n− ). Then −

an ] [a1 b1 ...δ bn ,

ǫa1 ...an ǫb1 ...bn

=

(−1)n n! δ

ǫa1 ...an ǫa1 b2 ...bn

=

(−1)n (n − 1)! δ

ǫa1 ...an ǫa1 a2 b3 ...bn

= .. .

ǫa1 ...an ǫa1 ...ar br+1 ...bn

=

−

−

an ] [a2 b2 ...δ bn ,

(−1)n 2(n − 2)! δ −

(1.11.5) (1.11.6)

an ] [a3 b3 ...δ bn ,

(−1)n r! (n − r)! δ

(1.11.7)

[ar+1 an ] br+1 ...δ bn .

(1.11.8)

Consider, for example, the case where n = 3 and n− = 0, i.e., where gab is positive definite. (The general case is handled similarly.) Then (1.11.5) comes out as the assertion: ǫabc ǫmnq = 6 δ

[a c] b m δ n δ q.

To see

that it holds, consider any anti-symmetric tensor αmnq at a point. Then αmnq = k ǫmnq for some k. So ǫabc ǫmnq αmnq

=

k ǫabc ǫmnq ǫmnq = 6 k ǫabc = 6 αabc

=

6 δ [am δ bn δ c]q αmnq .

Thus for all anti-symmetric αmnq at the point, we have ǫabc ǫmnq − 6 δ [am δ bn δ c]q αmnq = 0.

In particular, given arbitrary vectors λm , ρn , µq there,

ǫabc ǫmnq − 6 δ [am δ bn δ c]q λ[m ρn µq] = 0.

But since the expression in parentheses is itself anti-symmetric in the indices (m, n, q), this condition can be expressed as

ǫabc ǫmnq − 6 δ [am δ bn δ c]q λm ρn µq = 0.

Since λm , ρn , and µq are arbitrary, it follows that

ǫabc ǫmnq − 6 δ [am δ bn δ c]q = 0. This gives us (1.11.5). Next, (1.11.6) follows from (1.11.5) since ǫabc ǫanq

=

6 δ [aa δ bn δ c]q

=

2 δ aa δ [b n δ c] q − 2 δ [b n δ c] q

=

2(3 − 2)δ [b n δ c] q = 2 δ [b n δ c] q .

100


Finally, (1.11.7) follows from (1.11.6) since ǫabc ǫabq = 2 δ

[b

c] bδ q

= δ bb − 1 δ cq = 2 δ cq .

Another fact we shall need is ∇m ǫa1 ...an = 0

(1.11.9)

(where ∇ is the derivative operator on M determined by gab ). To see this, suppose λa is an arbitrary smooth field on M . Then, since λm ∇m ǫb1 ...bn is a smooth n-form on M , we have λm ∇m ǫb1 ...bn = ϕ ǫb1 ...bn for some scalar field ϕ. But then −

ϕ (−1)n n! = =

ϕ ǫa1 ...an ǫa1 ...an = ǫa1 ...an λm ∇m ǫa1 ...an 1 − 1 m λ ∇m ǫa1 ...an ǫa1 ...an = λm ∇m ((−1)n n!) = 0. 2 2

So ϕ = 0 and, hence, λm ∇m ǫb1 ...bn = 0. Since λm was arbitrary, we have (1.11.9).

Finally, we show how to recover ordinary vector analysis in terms of volume elements. Suppose our manifold M is R3 , gab is the Euclidean metric defined by equation (1.9.13), ∇ is the derivative operator determined by gab , and ǫabc is a volume element on M . Then, given contravariant vectors ξ and η at

some point, we define their dot and cross products as follows: ξ·η ξ×η

= ξ a ηa , = ǫabc ξb ηc .

(We are deliberately not using indices on the left.) It follows immediately from the anti-symmetry of ǫabc that ξ × η = −(η × ξ), and that ξ × η is orthogonal to both ξ and η. Furthermore, if we define the angular measure ∡(ξ, η) by setting

cos ∡(ξ, η) = 1

ξ·η , kξk kηk

where kξk = (ξ · ξ) 2 , then the magnitude of ξ × η is given by kξ × ηk

12

=

ǫabc ξb ηc ǫamn ξ m η n

=

2 δ [b m δ c] n ξ m η n ξb ηc

=

12

(ξ b ξb )(η c ηc ) − (ξ b ηb )2

12

1 = kξk kηk 1 − cos2 ∡(ξ, η) 2 = kξk kηk sin ∡(ξ, η). Consider an example. One learns in ordinary vector analysis that, given any three vectors α, β, γ at a point, γ × (α × β) = α(γ · β) − β(γ · α).

101


In our notation, this comes out as the assertion ǫabc γb ǫcmn αm β n = αa (γb β b ) − β a (γb αb ),

and it follows easily from equation (1.11.6): ǫabc γb ǫcmn αm β n

=

ǫcab ǫcmn γb αm β n

=

2 δ [am δ b]n γb αm β n = αa (γb β b ) − β a (αb γb ).

Given a smooth scalar field f and a smooth contravariant vector field ξ on M , we define the following: grad(f ) =

∇a f

div(ξ)

=

∇a ξ a

curl(ξ)

=

ǫabc ∇b ξc .

(In the more familiar notation usually found in textbooks, these would be written as ∇f , ∇ · ξ, and ∇ × ξ.) With these definitions, we can recover all the usual formulas of vector analysis. Here are two

simple examples. (Others are listed in the problems that follow.) (1) curl(grad f ) = 0. (2) div(curl ξ) = 0.

The first comes out as the assertion that ǫabc ∇b ∇c f = 0, which is immediate since ∇b ∇c f is symmetric in

(b, c). (For this result, flatness is not required.) The second comes out as ∇a (ǫabc ∇b ξc ) = 0. This follows

from equation (1.11.9) and the fact (now using flatness) that ∇a ∇b ξc is symmetric in (a, b).

Problem 1.11.1. One learns in the study of ordinary vector analysis that, for all vectors ξ, η, θ, and λ at a point, the following identities hold. (1) (ξ × η) · (θ × λ) = (ξ · θ)(η · λ) − (ξ · λ)(η · θ). (2) (ξ × (η × θ)) + (θ × (ξ × η)) + (η × (θ × ξ)) = 0. Reformulate these assertions in our notation and prove them. Problem 1.11.2. Do the same for the following assertion: div(ξ × η) = η · curl(ξ) − ξ · curl(η). (Here ξ and η are understood to be smooth vector fields.) Problem 1.11.3. We have seen (proposition 1.9.9) that every Killing field ξ a in n-dimensional Euclidean space (n ≥ 1) can be expressed uniquely in the form ξb = χa Fab + kb ,


102

where Fab and kb are constant, Fab is anti-symmetric, and χa is the position field relative to some point p. Consider the special case where n = 3. Let ǫabc be a volume element. Show that (in this special case) there is a unique constant field W a such that Fab = ǫabc W c . (If W a = 0, ξ a is the “infinitesimal generator” of a family of translations in the direction k a . Alternatively, if k a = 0, it generates a family of rotations about the point p with axis W a .)

(Hint: Consider W a = 12 ǫabc Fbc .)

Chapter 2

Classical Relativity Theory 2.1

Relativistic Spacetimes

With the basic ideas of differential geometry now at our disposal, we turn to relativity theory. It is helpful to think of the theory as determining a class of geometric models for the spacetime structure of our universe (and isolated subregions thereof, such as, for example, our solar system). Each represents a possible world (or world-region) compatible with the constraints of the theory. We describe these models in stages. First, we characterize a broad class of “relativistic spacetimes” and discuss their interpretation. Later, we introduce further restrictions involving global spacetime structure and Einstein’s equation. We take a relativistic spacetime to be a pair (M, gab ), where M is a smooth, connnected, fourdimensional manifold and gab is a smooth metric on M of Lorentz signature (1, 3). We interpret M as the manifold of point “events” in the world.1 The interpretation of gab is given by a network of interconnected physical principles. We list three in this section that are relatively simple in character because they make reference only to point particles and light rays. (These objects alone suffice to determine the metric, at least up to a constant.) We list a fourth in section 2.3 that concerns the behavior of (ideal) clocks. Still other principles involving generic matter fields will come up later. In what follows, let (M, gab ) be a fixed relativistic spacetime and let ∇ be the unique derivative

operator on M compatible with gab . Since gab has signature (1, 3), at every point p in M , the tangent 1 We

use “event” as a neutral term here and intend no special significance. Some might prefer to speak, for example, of

“equivalence classes of coincident point events” or “point event locations”. We shall take this interpretation for granted in what follows and shall, for example, refer to such things as “particle worldlines in a relativistic spacetime”.

103

104

CHAPTER 2. CLASSICAL RELATIVITY THEORY

1

4

space Mp has a basis ξ a , ..., ξ a such that, for all i and j in {1, 2, 3, 4},   +1 if i = 1 i i ξa ξa =  −1 if i = 2, 3, 4 1 1

2 2

3 3

Pn

1 1

2 2

3 3

4 4

j

i

and ξ a ξ a = 0 if i 6= j. It follows that given any vectors µa =

i=1

i

i

µ ξ a and ν a =

4 4

µa ν a = µ ν − µ ν − µ ν − µ ν

Pn

i=1

i

i

ν ξ a at p, (2.1.1)

and µa µa = µ µ − µ µ − µ µ − µ µ.

(2.1.2)

(Recall (1.9.3).) Given a vector η a at a point in M , we say η a is timelike

if η a ηa

>

0,

a

=

0,

a

≥

0,

null (or lightlike) if η ηa causal spacelike

if η ηa a

if η ηa

0. We claim first that ′ gab η a η b = k gab η a η b

(2.1.4)

for all η a at p. If η a is null with respect to both metrics, the assertion is trivial. So there are two cases to consider. Case 1: η a is timelike with respect to both metrics. Consider the following quadratic equation (in the variable x): 0 = gab (ξ a + x η a )(ξ b + x η b ) = gab ξ a ξ b + 2 x gab ξ a η b + x2 gab η a η b . The discriminant 4 (gab ξ a η b )2 − 4 (gab ξ a ξ b )(gab η a η b ) is positive (since (gab ξ a ξ b ) < 0, and (gab η a η b ) > 0). So the equation has real roots r1 and r2 with r1 · r2 =

gab ξ a ξ b . gab η a η b

(2.1.5)

Now the equation ′ 0 = gab (ξ a + x η a )(ξ b + x η b )

must have exactly the same roots as the preceding one (since the metrics agree on null vectors). So we also have r1 · r2 =

′ gab ξaξb ′ ηa ηb . gab

(2.1.6)

These two expressions for r1 · r2 , together with (2.1.3), yield (2.1.4). Case 2: η a is spacelike with respect to both metrics. Let γ a be any vector at p that is timelike with respect to both. Repeating the argument used for case 1, with η a now playing the role of ξ a , we have ′ gab η a η b gab ηa ηb = . ′ a b gab γ γ gab γ a γ b

(2.1.7)

′ But gab γ a γ b = k gab γ a γ b , because γ a falls under case 1. So η a must satisfy (2.1.4) in this case too. ′ Thus, we have established our claim. Since (gab − k gab ) is symmetric, it now follows by proposition 1.4.3 ′ that gab = k gab at p.

To complete the proof we define a scalar field Ω : M → R by setting Ω(p) =

p k(p) at each point p

′ ′ (where k(p) is determined as above). Then gab = Ω2 gab , and Ω is smooth since gab and gab are.

It turns out that dimension plays a role in proposition 2.1.1. Our spacetimes are four-dimensional. Suppose we temporarily drop that restriction and, for any n ≥ 2, consider “n-dimensional spacetimes”

(M, gab ) where M has dimension n and gab has signature (1, n − 1). What happens to the proposition?

The proof we have given carries over intact for all n ≥ 3. And even when n = 2, it carries over in part. Three versions of condition (1) are still equivalent to each other — those involving agreement on timelike,

108


causal, or spacelike vectors — and equivalent to condition (2). But in that special case, two metrics can agree on null vectors without being conformally equivalent. (At any point p in M , a “90 degree rotation” of Mp takes null vectors to null vectors, but it takes timelike vectors to spacelike vectors.) Problem 2.1.1. Consider our characterization of timelike vectors in terms of null vectors in the proof of proposition 2.1.1. Why does it fail if n = 2? Problem 2.1.2. (i) Show that it is possible to characterize timelike vectors (and so also null vectors and spacelike vectors) in terms of causal vectors. (ii) Show that it is possible to characterize timelike vectors (and so also null vectors and causal vectors) in terms of spacelike vectors. (Both characterizations should work for all n ≥ 2.) Conformally equivalent metrics do not agree, in general, on which curves qualify as geodesics or even just as geodesics up to reparametrization. But, it turns out, they do necessarily agree on which null curves are geodesics up to reparametrization. Indeed, we have the following proposition. Notice that clauses (1) and (2) correspond, respectively, to interpretive principles (C1) and (C2) above. Proposition 2.1.2. The following conditions are equivalent. ′ (1) gab and gab agree on which smooth curves on M are timelike. ′ and gab agree on which smooth curves on M can be reparameterized so as to be null geodesics. (2) gab ′ (3) gab and gab are conformally equivalent.

Proof. The implication (1) ⇒ (3) follows immediately from the preceding proposition. So does the impli-

cation (2) ⇒ (1). (Two metrics cannot agree on which curves are null geodesics up to reparametrization without first agreeing on which curves are null.) To complete the proof, we show that (3) implies (2). ′ ′ Assume that gab = Ω2 gab . Let γ be any smooth curve that is null (with respect to both gab and gab ),

and let λa be its tangent field. Further, let ∇′a be the unique derivative operator on M compatible with ′ gab . Then, by propositions 1.7.3 and 1.9.5,

λn ∇′n λa = λn (∇n λa − C anm λm ) where C anm = −

i 1 h a δ n ∇m Ω2 + δ am ∇n Ω2 − gnm g ar ∇r Ω2 . 2 2Ω

Substituting for C anm in the first equation, and using the fact that λa is null, we arrive at λn ∇′n λa = λn ∇n λa +

1 n (λ ∇n Ω2 ) λa . Ω2

It follows that λn ∇′n λa is everywhere proportional to λa iff λn ∇n λa is everywhere proportional to λa . Therefore, by proposition 1.7.9, γ can be reparametrized so as to be a geodesic with respect to gab iff it

′ can be so reparametrized with respect to gab .


109

Question: What would go wrong if we attempted to adapt the proof to show that conformally equivalent metrics agree as to which smooth timelike curves are geodesics up to reparametrization? We can understand the proposition to assert that the spacetime metric gab is determined up to a conformal factor, independently, by the set of possible worldlines of massive point particles and by the set of possible trajectories of light rays. ′ Next we turn to projective structure. Recall that gab is said to be projectively equivalent to gab if, for ′ all smooth curves γ on M , γ can be reparametrized so as to be geodesic with respect to gab iff it can be so

reparametrized with respect to gab . We have proved (proposition 1.9.6) that if the two metrics are both conformally and projectively equivalent, then the conformal factor connecting them is constant. Now, with interpretive principle P1 in mind, we prove a slightly strengthened version of the proposition that makes reference only to timelike geodesics (rather than arbitrary geodesics). To do so, we first strengthen proposition 1.4.3. ...ar Proposition 2.1.3. Let αba11...b be a tensor at some point in M . Suppose that s ...ar (1) αba11...b is symmetric in indices b1 , ..., bs , and s ...ar b1 (2) αba11...b ξ ... ξ bs = 0 for all timelike vectors ξ a at the point. s ...ar Then αba11...b = 0. s

Proof. Consider first the case where we are dealing with a tensor of form αb1 ...bs , i.e., one with no contravariant indices. Let ξ a be a timelike vector at the point in question, and let η a be an arbitrary vector there. Then there is an ǫ > 0 such that, for all real numbers x, if |x| < ǫ, (ξ a + x η a ) is timelike. Now consider the polynomial function f : R → R defined by f (x)

= αb1 ...bs (ξ b1 + x η b1 ) ... (ξ bs + x η bs ) s = αb1 ...bs ξ b1 ... ξ bs + x αb1 ...bs ξ b1 ... ξ bs−1 η bs + ... 1 s + xs−1 αb1 ...bs ξ b1 η b2 ... η bs + xs αb1 ...bs η b1 ... η bs . s−1

By our hypothesis, f (x) = 0 for all x in the interval (−ǫ, ǫ). Hence all derivatives of f vanish in the interval. So αb1 ...bs η b1 ... η bs = 0. Since η a was an arbitrary vector at our point, it follows, by proposition 1.4.3, that αb1 ...bs = 0 there. For the general case, let µa1 ...νar be arbitrary vectors at the ...ar point. Then αba11...b µa1 ...νar = 0 by the argument just given. So (since µa1 ...νar are arbitrary vectors), s ...ar αba11...b = 0. s ...ar Of course, a parallel proposition holds if αba11...b is symmetric in indices a1 , ..., ar . Indeed, we can arrive s

at that formulation simply by lowering the a-indices and raising the b-indices, applying the proposition as proved, and then restoring the original index positions. Problem 2.1.3. Does proposition 2.1.3 still hold if condition (1) is left intact but (2) is replaced by


110

...ar b1 (2 ′ ) αba11...b ξ ... ξ bs = 0 for all spacelike vectors ξ a at the point? s

And what if it is replaced by ...ar b1 (2 ′′ ) αba11...b ξ ... ξ bs = 0 for all null vectors ξ a at the point? s

Justify your answers. The proposition we are after is the following. ′ ′ Proposition 2.1.4. Assume gab = Ω2 gab . Further, assume gab and gab agree as to which smooth, timelike

curves can be reparametrized so as to be geodesics. Then Ω is constant. ′ Proof. Assume ∇′ = (∇, C abc ) where, once again, ∇′ is the derivative operator associated with gab . It

suffices for us to show that C abc = δ ab ϕc + δ ac ϕb for some smooth field ϕa . For then the constancy of Ω follows exactly as in our proof of proposition 1.9.6. To show that C abc has this form, we need only make a slight revision in our proof of proposition 1.7.10. There we started from the assumption that ∇′ and ∇ agree as to which (arbitrary) smooth

curves can be reparametrized so as to be geodesics. Using that assumption, we showed that the field ϕadbcr = C abc δ dr − C dbc δ ar satisfies the condition ϕad(bcr) ξ b ξ c ξ r = 0

(2.1.8)

for all vectors ξ a at all points. Then we invoked proposition 1.4.3 to conclude that ϕad(bcr) = 0 everywhere. Arguing in exactly the same way from our weaker assumption (that the metrics agree as to which smooth, timelike curves can reparametrized so as to be geodesics), we can show that (2.1.8) holds for all timelike vectors at all points. But we know (by proposition 2.1.3) that this condition also forces the conclusion that ϕad(bcr) = 0 everywhere. The rest of the proof goes through exactly as in that of proposition 1.7.10. Without reference to particular types of vectors, we can show that C abc = δ ab ϕc + δ ac ϕb where 1 ϕc = Cd . n + 1 cd Later in this book we shall consider a few (not many) particular examples of spacetimes. But one should be mentioned immediately, namely Minkowski spacetime. We take it to be the pair (M, gab ) where (i) M is the manifold R4 , (ii) (M, gab ) is flat, i.e., has vanishing Riemann curvature everywhere, and (iii) (M, gab ) is geodesically complete, i.e., all maximally extended geodesics have domain R. Minkowski spacetime is very special because its structure as an affine manifold (M, ∇) is precisely the

same as that of four-dimensional Euclidean space. (Here, of course, ∇ is understood to be the unique

derivative operator on M compatible with gab .) In particular, given any point o in M , there is a smooth “direction field” χa on M that vanishes at o and satisfies the condition ∇a χb = δab . (Recall proposition

1.7.12.)

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2.2

Temporal Orientation and “Causal Connectibility”

The characterization we have given of relativistic spacetimes is extremely loose. Many further conditions might be imposed. We consider one in this section, namely “temporal orientability”. First we need to review certain basic facts about Lorentzian metrics. Once again, let (M, gab ) be a fixed relativistic spacetime. We start with the orthogonality relation that gab determines in the tangent space at every point of M . (Two vectors µa and ν a at a point qualify as orthogonal, of course, if µa νa = 0.) Proposition 2.2.1. Let µa and ν a be vectors at some point p in M . Then the following both hold. (1) If µa is timelike and ν a is orthogonal to µa , then either ν a = 0 or ν a is spacelike. (2) If µa and ν a are both null, then they are orthogonal iff they are proportional (i.e., one is a scalar multiple of the other). 1

4

1

1

i

i

Proof. (1) Let ξ a , ..., ξ a be an orthonormal basis for Mp with ξ a ξ a = 1, and ξ a ξ a = −1 for i = 2, 3, 4. Pn i i a Pn i i a a Then we can express µa and ν a in the form µ a = and ν a = i=1 µ ξ i=1 ν ξ . Now assume µ

is timelike, ν a is orthogonal to µa , and ν a 6= 0. We show that ν a is spacelike. It follows from our

assumptions that

1

(µ)2

2

1 1

µν

3

4

(µ)2 + (µ)2 + (µ)2 ,

>

2 2

=

3 3

(2.2.1)

4 4

µ ν + µ ν + µ ν,

(2.2.2)

1

2

3

4

µ 6= 0,

(ν)2 + (ν)2 + (ν)2

(2.2.3)

> 0.

(2.2.4)

(The first two assertions follow from (2.1.2) and (2.1.1). The third follows from the first. For the final 2

3

4

2

3

4

1

inequality, note that if (ν)2 + (ν)2 + (ν)2 = 0, then ν = ν = ν = 0, and so, by (2.2.2) and (2.2.3), ν = 0 as well. This contradicts our assumption that ν a 6= 0.) In turn, it now follows by the Schwarz inequality 2

3

4

2

3

4

(as applied to the vectors (µ, µ, µ) and (ν, ν, ν)) that 1

1

2 2

3 3

4 4

2

3

4

2

3

4

1

2

3

4

(µ)2 (ν)2 = (µ ν + µ ν + µ ν)2 ≤ [(µ)2 + (µ)2 + (µ)2 ] [(ν)2 + (ν)2 + (ν)2 ] < (µ)2 [(ν)2 + (ν)2 + (ν)2 ], and hence, by (2.2.3) again, that 1

2

3

4

(ν)2 < (ν)2 + (ν)2 + (ν)2 . Thus ν a is spacelike. (2) Assume µa and ν a are both null. If they are proportional, then they are trivially orthogonal. For if, say, µa = k ν a , then µa νa = k(ν a νa ) = 0 (since ν a is null). Assume, conversely, that the vectors are orthogonal. Let ξ a be a timelike vector at p. By clause (1) — since ν a is not spacelike — either ν a = 0 or ξ a νa 6= 0. (Here ξ a is playing role of µa .) In the first case, µa and ν a are trivially proportional. So we may assume that ξ a νa 6= 0. Then there is a number k such that k (ξ a νa ) = ξ a µa . Hence, (µa − k ν a ) ξa = 0.

Now (µa − k ν a ) is not spacelike. (The right side of

(µa − k ν a )(µa − k νa ) = µa µa − 2 k (µa νa ) + k 2 (ν a νa )


112

is 0 since, by assumption, µa and ν a are null and µa νa = 0.) So, by clause (1) again, it must be the case that (µa − k ν a ) = 0, i.e., µa and ν a are proportional. Problem 2.2.1. Let p be a point in M . Let p be a point in M . Show that there is no two-dimensional subspace of Mp all of whose elements are causal (timelike or null). Problem 2.2.2. Let g ′ab be a second metric on M (not necessarily of Lorentz signature). Show that the following conditions are equivalent. (1) For all p in M , gab and g ′ab agree on which vectors at p are orthogonal. (2) g ′ab is conformally equivalent to either gab or −gab . Next we consider the “lobes” of the null cone determined by gab at points of M . Let us say that two timelike vectors µa and ν a at a point are co-oriented (or have the same orientation) if µa νa > 0. Proposition 2.2.2. For all points p in M , co-orientation is an equivalence relation on the set of timelike vectors in Mp . Proof. Reflexivity and symmetry are immediate. For transitivity, let µa , ν a , ω a be timelike vectors at a point, with the pairs {µa , ν a } and {ν a , ω a } both co-oriented. We must show that {µa , ω a } is co-oriented as well. The argument is very much like that for the second clause of proposition 2.2.1.

Since µa νa > 0 and ω a νa > 0, there is a real number k > 0 such that µa νa = k (ω a νa ). Hence, (µa − k ω a )νa = 0. Since ν a is timelike, we know from the first clause of proposition 2.2.1 that either

(µa − k ω a ) is the zero-vector 0 or it is spacelike. In the first case, µa = k ω a , and so the pair {µa , ω a } is

certainly co-oriented (µa ωa = k (ω a ωa ) > 0). So we may assume that (µa − k ω a ) is spacelike. But then µa µa − 2 k (µa ωa ) + k 2 (ω a ωa ) = (µa − k ω a )(µa − k ωa ) < 0.

Since µa µa , ω a ωa , and k are all positive, it follows that µa ωa is positive as well. So, again, we are led to the conclusion that the pair {µa , ω a } is co-oriented. The equivalence classes determined at each point by the co-orientation relation will be called temporal lobes. There must be at least two lobes at each point since, for any timelike vector µa there, µa and −µa are not co-oriented. There cannot be more than two since, for all timelike µa and ν a at a point, ν a is co-oriented either with µa or with −µa . (Remember, two timelike vectors at a point cannot be

orthogonal.) Hence there are exactly two lobes at each point. It is easy to check that each lobe is convex, i.e., if µa and ν a are co-oriented at a point, and a, b are both positive real numbers, then (a µa + b ν a ) is a timelike vector at the point that is co-oriented with µa and ν a . The relation of co-orientation can be extended easily to the larger set of non-zero causal (i.e., timelike or null) vectors. Given any two such vectors µa and ν a at a point, we can take them to be co-oriented if either µa νa > 0 or ν a = k µa with k > 0. (The second possibility must be allowed since we want a


113

non-zero null vector to count as being co-oriented with itself.) Once again, co-orientation turns out to be an equivalence relation with two equivalence classes that we call causal lobes. (Only minor changes in the proof of proposition 2.2.2 are required to establish that the extended co-orientation relation is transitive.) These lobes, too, are convex. For future reference, we record two more facts about Lorentz metrics. (Let us agree to write kµa k for 1

1

(µa µa ) 2 when µa is causal, and write it for (−µa µa ) 2 when µa is spacelike.)

Proposition 2.2.3. Let µa and ν a be causal vectors at some point p in M . Then the following both hold. (1) (“Wrong way Schwarz inequality”) |µa νa | ≥ kµa k kν a k, with equality iff µa and ν a are proportional.

(2) (“Wrong way triangle inequality”) If µa and ν a are non-zero and co-oriented, kµa + ν a k ≥ kµa k + kν a k, with equality iff µa and ν a are proportional. Proof. (1) If both µa and ν a are null, the assertion follows immediately from the second assertion in proposition 2.2.1. So we may assume that one of the vectors, say µa , is timelike. Now we can certainly express ν a in the form ν a = k µa + σ a , with k a real number and σ a a vector at p orthogonal to µa . (It suffices to take k = (µa νa )/(µa µa ) and σ a = (ν a − k µa ).) Hence, µa νa

=

k (µa µa ),

ν a νa

=

k 2 (µa µa ) + σ a σa .

Since σ a is orthogonal to µa , it must either be spacelike or the zero vector (by proposition 2.2.1). In either case, (σ a σa ) ≤ 0. So, since (µa µa ) > 0 and (ν a νa ) ≥ 0, it follows that (µa νa )2 = k 2 (µa µa )2 = [(ν a νa ) − (σ a σa )] (µa µa ) ≥ (ν a νa ) (µa µa ) = kµa k2 kν a k2 . Equality holds here iff (σ a σa ) = 0. But (as noted already), σ a is either the zero vector or spacelike (in which case (σ a σa ) < 0). So equality holds iff σ a = 0, i.e., ν a = k µa . We leave the second clause as an exercise. Problem 2.2.3. Prove the second clause of proposition 2.2.3. Now we switch our attention to considerations of global null cone structure. We say that (M, gab ) is temporally orientable if there exists a continuous timelike vector field τ a on M . Suppose the condition is satisfied. Then we take two such fields τ a and τ ′a to be co-oriented if they are so at every point, i.e., if τ a τ ′ a > 0 holds at every point of M . Co-orientation, now understood as a relation on continuous timelike vector fields, is an equivalence relation with two equivalence classes. (It inherits this property from the original relation defined on timelike vectors at individual points.) A temporal orientation of (M, gab ) is a

114


choice of one of those two equivalence classes to count as the “future” one. Thus, a non-zero causal vector ξ a at a point of M is said to be future-directed or past-directed with respect to the temporal orientation T depending on whether τ a ξa > 0 or τ a ξa < 0 at the point, where τ a is any continuous timelike vector

field in T . (Remember, τ a ξa cannot be 0, since no timelike vector can be orthogonal to a non-zero causal

vector.) Derivatively, a smooth, causal curve γ: I → M is said to be future directed (resp. past directed )

with respect to T if its tangent vector at every point is so.

Our characterization of “relativistic spacetimes” in the preceding section does not guarantee temporal orientability. But we shall take the condition for granted in what follows. We assume that our background spacetime (M, gab ) is temporally orientable and that a particular temporal orientation has been specified. Also, given points p and q in M , we shall write p ≪ q (resp. p < q) if there is a smooth, future-

directed, timelike (resp. causal) curve γ : [a, b] → M where γ(a) = p and γ(b) = q. Note that p < p, for

all points p in all spacetimes. (This is the case because the zero vector in the tangent space at any point qualifies as a null vector.) But it is not the case, in general, that p ≪ p. The latter condition holds iff

there is a smoooth, closed, future-directed timelike curve that begins and ends at p. The two relations ≪ and < are naturally construed as relations of “causal connectibility (or accessibility)”. Appendix: Recovering Geometric Structure from the Causal Connectibility Relation

We started with a spacetime model (M, gab ) exhibiting several levels of geometric structure, and used the latter to define the relations ≪ and < on M .7 The question now arises whether it is possible to work

backwards, i.e., start with the pair (M, ≪) or (M, 0, there exists a

smooth, future-directed timelike curve γ from p to q with kγk < ǫ. (But there is no such curve with length

0, since all timelike curves have non-zero length.)

Though some work is required to give the proposition an honest proof (see O’Neill [46, pp. 294-5]), it should seem intuitively plausible. If there is a smooth, timelike curve connecting p and q, there is also a jointed, zig-zag null curve connecting them. It has length 0. But we can approximate the jointed null q short timelike curve long timelike curve

p

Figure 2.3.1: A long timelike curve from p to q and a very short one that approximates a broken null curve.

120


curve arbitrarily closely with smooth timelike curves that swing back and forth. So (by the continuity of the length function), we should expect that, for all ǫ > 0, there is an approximating timelike curve that has length less than ǫ. (See figure 2.3.1.) The answer to the second question (“Can one maximize recorded elapsed time between p and q?”) is “yes” if one restricts attention to local regions of spacetime. In the case of positive definite metrics, i.e., ones with signature of form (n, 0), we know geodesics are “locally shortest” curves. The corresponding result for Lorentzian metrics is that timelike geodesics are “locally longest” curves. Proposition 2.3.3. Let γ : I → M be a smooth, future-directed, timelike curve. Then γ can be repara-

metrized so as to be a geodesic iff for all s ∈ I, there exists an open set O containing γ(s) such that,

for all s1 , s2 ∈ I with s1 ≤ s ≤ s2 , if the image of γ ′ = γ|[s1 ,s2 ] is contained in O, then γ ′ (and its

reparametrizations) are longer than all other timelike curves in O from γ(s1 ) to γ(s2 ). (Here γ|[s1 ,s2 ] is the restriction of γ to the interval [s1 , s2 ].)

The proof of the proposition is very much the same as in the positive definite case. (See Hawking and Ellis [30, p. 105].) Thus, of all clocks passing locally from p to q, the one that will record the greatest elapsed time is the one that “falls freely” from p to q. To get a clock to read a smaller elapsed time than the maximal value one will have to accelerate the clock. Now, acceleration requires fuel, and fuel is not free. So proposition 2.3.3 has the consequence that (locally) “saving time costs money”. And proposition 2.3.2 may be taken to imply that “with enough money one can save as much time as one wants”.

γ2

γ1 γ3

Figure 2.3.2: Two-dimensional Minkowski spacetime rolled-up into a cylindrical spacetime. Three timelike curves are displayed: γ1 and γ3 are geodesics; γ2 is not; γ1 is longer than γ2 ; and γ2 is longer than γ3 . The restriction here to local regions of spacetime is essential. The connection described between clock behavior and acceleration does not, in general, hold on a global scale. In some relativistic spacetimes,


121

one can find future-directed timelike geodesics connecting two events that have different lengths, and so clocks following the curves will record different elapsed times between the events even though both are in a state of free fall. Furthermore — this follows from the preceding claim by continuity considerations alone — it can be the case that of two clocks passing between the events, the one that undergoes acceleration during the trip records a greater elapsed time than the one that remains in a state of free fall. (A rolled up version of two-dimensional Minkowski spacetime provides a simple example. See figure 2.3.2.) The connection we have been considering between clock behavior and acceleration was once thought to be paradoxical. Recall the so-called “clock paradox”. Suppose two clocks, A and B, pass from one event to another in a suitably small region of spacetime. Further suppose A does so in a state of free fall but B undergoes acceleration at some point along the way. Then, we know, A will record a greater elapsed time for the trip than B. This was thought paradoxical because it was believed that relativity theory denies the possibility of distinguishing “absolutely” between free fall motion and accelerated motion. (If we are equally well entitled to think that it is clock B that is in a state of free fall and A that undergoes acceleration, then, by parity of reasoning, it should be B that records the greater elapsed time.) The resolution of the paradox, if one can call it that, is that relativity theory makes no such denial. The situations of A and B here are not symmetric. The distinction between accelerated motion and free fall makes every bit as much sense in relativity theory as it does in Newtonian physics. In what follows, unless indication is given to the contrary, a “timelike curve” should be understood to be a smooth, future-directed, timelike curve parametrized by elapsed proper time, i.e., by arc length. In that case, the tangent field ξ a of the curve has unit length (ξ a ξa = 1). And if a particle happens to have the image of the curve as its worldline, then, at any point, ξ a is called the particle’s four-velocity there.

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2.4

Space/Time Decomposition at a Point and Particle Dynamics

Let γ be a smooth, future-directed, timelike curve with unit tangent field ξ a in our background spacetime (M, gab ). We suppose that some massive point particle O has (the image of) this curve as its worldline. Further, let p be a point on the image of γ and let λa be a vector at p. Then there is a natural decomposition of λa into components proportional to, and orthogonal to, ξ a : λa =

(λb ξb )ξ a | {z }

+

proportional to ξ a

(λa − (λb ξb )ξ a ) . | {z }

(2.4.1)

orthogonal to ξa

These are standardly interpreted, respectively, as the “temporal” and “spatial” components of λa relative to ξ a (or relative to O). In particular, the three-dimensional vector space of vectors at p orthogonal to ξ a is interpreted as the “infinitesimal” simultaneity slice of O at p.11 If we introduce the tangent and orthogonal projection operators kab

=

ξa ξb ,

(2.4.2)

hab

=

gab − ξa ξb ,

(2.4.3)

then the decomposition can be expressed in the form λa = k ab λb + hab λb .

(2.4.4)

We can think of kab and hab as the relative temporal and spatial metrics determined by ξ a . They are symmetric and satisfy k ab k bc

= k ac ,

(2.4.5)

hab hbc

= hac .

(2.4.6)

Many standard textbook assertions concerning the kinematics and dynamics of point particles can be recovered using these decomposition formulas. For example, suppose that the worldline of a second particle O′ also passes through p and that its four-velocity at p is ξ ′ a . (Since ξ a and ξ ′ a are both futuredirected, they are co-oriented, i.e., ξ a ξ ′ a > 0.) We compute the speed of O′ as determined by O. To do so, we take the spatial magnitude of ξ ′ a relative to O and divide by its temporal magnitude relative to O:12 v = speed of O′ relative to O = 1

khab ξ ′ b k . kk ab ξ ′ b k

(2.4.7) 1

(Recall that for any vector µa , kµa k is (µa µa ) 2 if µa is causal, and it is (−µa µa ) 2 otherwise.) From (2.4.2), (2.4.3), (2.4.5), and (2.4.6), we have

1

1

kk ab ξ ′ b k = (k ab ξ ′ b kac ξ ′ c ) 2 = (kbc ξ ′ b ξ ′ c ) 2 = (ξ ′ b ξb ) 11 Here

we simply take for granted the standard identification of “relative simultaneity” with orthogonality. For discussion

of how the identification is justified, see Malament [42, section 3.1] and further references cited there. 12 We

are, in effect, choosing units in which c = 1.

123


and 1

1

1

khab ξ ′ b k = (−hab ξ ′ b hac ξ ′ c ) 2 = (−hbc ξ ′ b ξ ′ c ) 2 = ((ξ ′ b ξb )2 − 1) 2 . So

1

v=

((ξ ′ b ξb )2 − 1) 2 < 1. (ξ ′ b ξb )

(2.4.8)

Thus, as measured by O, no massive particle can ever attain the maximal speed 1. (A similar calculation shows that, as determined by O, light always travels with speed 1.) For future reference, we note that (2.4.8) implies that

1 . (ξ ′ b ξb ) = √ 1 − v2

(2.4.9)

It is a basic fact of relativistic life that there is associated with every point particle, at every event on its worldline, a four-momentum (or energy-momentum) vector P a that is tangent to its worldline there. The length kP a k of this vector is what we would otherwise call the mass (or inertial mass or rest

mass) of the particle. So, in particular, if P a is timelike, we can write it in the form P a = m ξ a , where m = kP a k > 0 and ξ a is the four-velocity of the particle. No such decomposition is possible when P a is null and m = kP a k = 0.

Suppose a particle O with positive mass has four-velocity ξ a at a point, and another particle O′ has four-momentum P a there. The latter can either be a particle with positive mass or mass 0. We can recover the usual expressions for the energy and three-momentum of the second particle relative to O if we decompose P a in terms of ξ a . By (2.4.4) and (2.4.2), we have P a = (P b ξb ) ξ a + | {z } energy

hab P b } | {z

.

(2.4.10)

three−momentum

The energy relative to O is the coefficient in the first term: E = P b ξb . If O′ has positive mass and P a = m ξ ′ a , this yields, by (2.4.9), m E = m (ξ ′ b ξb ) = √ . 1 − v2

(2.4.11)

(If we had not chosen units in which c = 1, the numerator in the final expression would have been mc2 p and the denominator 1 − (v 2 /c2 ).) The three-momentum relative to O is the second term hab P b in the

decomposition of P a , i.e., the component of P a orthogonal to ξ a . It follows from (2.4.8) and (2.4.9) that it has magnitude

mv 1 . p = khab m ξ ′ b k = m ((ξ ′ b ξb )2 − 1) 2 = √ 1 − v2

(2.4.12)

Interpretive principle P1 asserts that the worldlines of free particles with positive mass are the images of timelike geodesics. It can be thought of as a relativistic version of Newton’s first law of motion. Now we consider acceleration and a relativistic version of the second law. Once again, let γ : I → M be a

smooth, future-directed, timelike curve with unit tangent field ξ a . Just as we understand ξ a to be the

four-velocity field of a massive point particle (that has the image of γ as its worldline), so we understand


124

ξ n ∇n ξ a — the directional derivative of ξ a in the direction ξ a — to be its four-acceleration field (or just

acceleration f ield). The four-acceleration vector at any point is orthogonal to ξ a . (This is clear, since

ξ a (ξ n ∇n ξa ) =

1 2

ξ n ∇n (ξ a ξa ) =

1 2

ξ n ∇n (1) = 0.) The magnitude kξ n ∇n ξ a k of the four-acceleration

vector at a point is just what we would otherwise describe as the curvature of γ there. It is a measure

of the rate at which γ “changes direction”. (And γ is a geodesic precisely if its curvature vanishes everywhere.) The notion of spacetime acceleration requires attention. Consider an example. Suppose you decide to end it all and jump off the Empire State Building. What would your acceleration history be like during your final moments? One is accustomed in such cases to think in terms of acceleration relative to the earth. So one would say that you undergo acceleration between the time of your jump and your calamitous arrival. But on the present account, that description has things backwards. Between jump and arrival, you are not accelerating. You are in a state of free fall and moving (approximately) along a spacetime geodesic. But before the jump, and after the arrival, you are accelerating. The floor of the observation deck, and then later the sidewalk, push you away from a geodesic path. The all-important idea here is that we are incorporating the “gravitational field” into the geometric structure of spacetime, and particles traverse geodesics if and only if they are acted on by no forces “except gravity”. The acceleration of our massive point particle, i.e., its deviation from a geodesic trajectory, is determined by the forces acting on it (other than “gravity”). If it has mass m, and if the vector field F a on I represents the vector sum of the various (non-gravitational) forces acting on it, then the particle’s four-acceleration ξ n ∇n ξ a satisfies

F a = m ξ n ∇n ξ a .

(2.4.13)

This is our version of Newton’s second law of motion. Consider an example. (Here we anticipate our discussion in section 2.6.) Electromagnetic fields are represented by smooth, anti-symmetric fields Fab . If a particle with mass m > 0, charge q, and fourvelocity field ξ a is present, the force exerted by the field on the particle at a point is given by q F ab ξ b . If we use this expression for the left side of (2.4.13), we arrive at the Lorentz law of motion for charged particles in the presence of an electromagnetic field: q F ab ξ b = m ξ b ∇b ξ a .

(2.4.14)

(Notice that the equation makes geometric sense. The acceleration field on the right is orthogonal to ξ a . But so is the force field on the left, since ξa (F ab ξ b ) = ξ a ξ b Fab = ξ a ξ b F(ab) , and F(ab) = 0 by the anti-symmetry of Fab .)

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2.5

The Energy-Momentum Field Tab

In classical relativity theory, one generally takes for granted that all that there is, and all that happens, can be described in terms of various “matter fields”, each of which is represented by one or more smooth tensor (or spinor) fields on the spacetime manifold M .13 The latter are assumed to satisfy particular “field equations” involving the spacetime metric gab . For present purposes, the most important basic assumption about the matter fields is the following. Associated with each matter field F is a symmetric smooth tensor field Tab characterized by

the property that, for all points p in M , and all future-directed, unit timelike vectors ξ a at p, T a b ξ b is the four-momentum density of F at p as determined relative to ξ a . Tab is called the energy-momentum field associated with F . The four-momentum density vector T a b ξ b

at a point can be further decomposed into its temporal and spatial components relative to ξ a , T a b ξ b = (Tmb ξ m ξ b ) ξ a + {z } | energy density

Tmb hma ξ b | {z }

,

three−momentum density

just as the four-momentum P a of a particle was decomposed in (2.4.10). The coefficient of ξ a in the first component, Tab ξ a ξ b , is the energy density of F at the point as determined relative to ξ a . The second

component, Tnb (g an − ξ a ξ n ) ξ b , is the three-momentum density of F there as determined relative to ξ a .

A number of assumptions about matter fields can be captured as constraints on the energy-momentum tensor fields with which they are associated. Examples are the following. (Suppose Tab is associated with matter field F .) Weak Energy Condition (WEC): Given any timelike vector ξ a at any point in M , Tab ξ a ξ b ≥ 0. Dominant Energy Condition (DEC): Given any timelike vector ξ a at any point in M , Tab ξ a ξ b ≥ 0 and T ab ξ b is timelike or null.

Strengthened Dominant Energy Condition (SDEC): Given any timelike vector ξ a at any point in M , Tab ξ a ξ b ≥ 0 and, if Tab 6= 0 there, then T ab ξ b is timelike. Conservation Condition (CC): ∇a T ab = 0 at all points in M. The weak energy condition asserts that the energy density of F , as determined by any observer at any

point, is non-negative. The dominant energy condition adds the requirement that the four-momentum density of F , as determined by any observer at any point, is a future-directed causal (i.e., timelike or null)

vector. We can understand this second clause to assert that the energy of F does not propagate with 13 This

being the case, the question arises as to how (or whether) one can adequately recover talk about “point particles”

in terms of the matter fields. We shall briefly discuss the question later in this section.

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superluminal velocity. The strengthened version of the dominant energy condition just changes “causal” to “timelike” in the second clause. It captures something of the flavor of (C1) in section 2.1, but avoids reference to “point particles”. Each of the listed energy conditions is strictly stronger than the ones that precede it (see problem 2.5.1). Problem 2.5.1. Give examples of each of the following. (1) A smooth symmetric field Tab that does not satisfy the WEC (2) A smooth symmetric field Tab that satisfies the WEC but not the DEC (3) A smooth symmetric field Tab that satisfies the DEC but not the SDEC Problem 2.5.2. Show that the DEC holds iff given any two co-oriented timelike vectors ξ a and η a at a point in M , Tab ξ a η b ≥ 0. The conservation condition, finally, asserts that the energy-momentum carried by F is locally con-

served. If two or more matter fields are present in the same region of spacetime, it need not be the

case that each one individually satisfies the condition. Interaction may occur. But it is a fundamental assumption that the composite energy-momentum field formed by taking the sum of the individual ones satisfies it. Energy-momentum can be transferred from one matter field to another, but it cannot be created or destroyed. The stated conditions have a number of consequences that support the interpretations just given. We mention two. The first requires a few preliminary definitions. A subset S of M is said to be achronal if there do not exist points p and q in S such that p ≪ q. Let

γ : I → M be a smooth curve. We say that a point p in M is a future-endpoint of γ if, for all open sets

O containing p, there exists an s0 in I such that, for all s ∈ I, if s ≥ s0 , then γ(s) ∈ O, i.e., γ eventually enters and remains in O. (Past-endpoints are defined similarly.) Now let S be an achronal subset of M .

The domain of dependence D(S) of S is the set of all points p in M with this property: given any smooth causal curve without (past- or future-) endpoint, if its image contains p, then it intersects S. (See figure 2.5.1.) So, in particular, S ⊆ D(S).

S

} D(S)

Figure 2.5.1: The domain of dependence D(S) of an achronal set S.


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In section 2.10, we shall make precise a sense in which “what happens on S determines what happens throughout D(S)”. Here we consider just one aspect of that determination. Proposition 2.5.1. Let S be an achronal subset of M . Further, let Tab be a smooth, symmetric field on M that satisfies both the dominant energy and conservation conditions. Finally, assume Tab = 0 on S. Then Tab = 0 on all of D(S). The intended interpretation of the proposition is clear. If energy-momentum cannot propagate (locally) outside the null-cone, and if it is conserved, and if it vanishes on S, then it must vanish throughout D(S). After all, how could it “get to” any point in D(S)? Note that our formulation of the proposition does not presuppose any particular physical interpretation of the symmetric field Tab . All that is required is that it satisfy the two stated conditions. (For a proof, see Hawking and Ellis [30, p. 94].) Now recall (P1). It asserts that free massive point particles traverse (images of) timelike geodesics. The next proposition (Geroch and Jang [24]) shows that it is possible, in a sense, to capture the principle as a theorem in relativity theory. The trick is to find a way to talk about “massive point particles” in the language of extended matter fields. In effect, we shall model them as nested sequences of small, but extended, bodies that converge to a point. It turns out that if the energy-momentum content of each body in the sequence satisfies appropriate conditions, then the convergence point will necessarily traverse (the image of) a timelike geodesic. Proposition 2.5.2. Let γ : I → M be smooth curve. Suppose that given any open subset O of M

containing γ[I], there exists a smooth symmetric field Tab on M such that the following conditions hold. (1) Tab satisfies the strengthened dominant energy condition. (2) Tab satisfies the conservation condition. (3) Tab = 0 outside of O. (4) Tab 6= 0 at some point in O.

Then γ is timelike and can be reparametrized so as to be a geodesic. The proposition might be paraphrased this way. Suppose that for some smooth curve γ, arbitrarily small bodies with energy-momentum satisfying conditions (1) and (2) can contain the image of γ in their worldtubes. Then γ must be a timelike geodesic (up to reparametrization). Bodies here are understood to be “free” if their internal energy-momentum is conserved (by itself). If a body is acted on by a field, it is only the composite energy-momentum of the body and field together that is conserved. Note that our formulation of the proposition takes for granted that we can keep the background spacetime metric gab fixed while altering the fields Tab that live on M . This is justifiable only to the extent that we are dealing with test bodies whose effect on the background spacetime structure is negligible.14 14 Stronger

theorems have been proved (see Ehlers and Geroch [16]) where one still models a point particle as a nested

sequence of extended bodies converging to a point but does not require that the perturbative effect of each body in the sequence disappear entirely. One requires only that, in a certain precise sense, it disappear in the limit.


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Note also that we do not have to assume at the outset that the curve γ is timelike. That follows from the other assumptions. We have here a precise proposition in the language of matter fields that, at least to some degree, captures (P1). Similarly, it is possible to capture (C2), concerning the behavior of light, with a proposition about the behavior of solutions to Maxwell’s equations in a limiting regime (“the optical limit”) where wavelengths are small. It asserts, in effect, that when one passes to this limit, packets of electromagnetic waves are constrained to move along (images of) null geodesics. (See Wald [60, p. 71].) It is worth noting that the Geroch-Jang result fails if condition (1) is dropped. Consider again our nested sequence of bodies converging to a point. It turns out that the conservation condition alone imposes no restrictions whatsoever on the wordline of the convergence point. It can be a null or spacelike curve. It can also be a timelike curve that exhibits any desired pattern of large and/or changing acceleration. The next proposition, based on a suggestion of Robert Geroch (in personal communication), gives a counterexample.15 Proposition 2.5.3. Let (M, gab ) be Minkowski spacetime, and let γ : I → M be any smooth timelike curve. Then, given any open subset O of M containing γ[I], there exists a smooth symmetric field Tab

on M that satisfies conditions (2), (3), and (4) in the preceding proposition. (If we want, we can also strengthen condition (4) and require that Tab be non-vanishing throughout some open subset O1 ⊆ O

containing γ[I].)

Proof. Let O be an open subset of M containing γ[I], and let f : M → R be any smooth scalar field on M . (Later we shall impose further restrictions on f .) Consider the fields S abcd = f (g ad g bc − g ac g bd )

and T ac = ∇b ∇d S abcd , where ∇ is the (flat) derivative operator on M compatible with gab . We have T ac = (g ad g bc − g ac g bd ) ∇b ∇d f = ∇c ∇a f − g ac (∇b ∇b f ).

(2.5.1)

So T ac is clearly symmetric. It is also divergence free since ∇a T ac = ∇a ∇c ∇a f − ∇c ∇b ∇b f = ∇c ∇a ∇a f − ∇c ∇b ∇b f = 0. (The second equality follows from the fact that ∇ is flat, and so ∇a and ∇c commute in their action on

arbitrary tensor fields.)

To complete the proof, we now impose further restrictions on f to insure that conditions (3) and (4) are satisfied. Let O1 be any open subset of M such that γ[I] ⊆ O1 and cl(O1 ) ⊆ O. (Here cl(A) is the

closure of A.) Our strategy will be to choose a particular f on O1 and a particular f on M −cl(O), and

then fill in the buffer zone O−cl(O1 ) any way whatsoever (so long as the resultant field is smooth). On M −cl(O), we simply take f = 0. This choice guarantees that, no matter how we smoothly extend f to

all of M , T ac will vanish outside of O. 15 It

is formulated in terms of an initial curve that is timelike — the case of greatest interest — but that is not essential.

The example can also be adapted to show that proposition 2.5.1 fails if the energy condition there is dropped.


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For the other specification, let o be any point in M and let χa be the “position field” on M determined relative to o. So ∇a χb = δa b everywhere, and χa = 0 at o. On O1 , we take f = −(χn χn ). With that

choice, T ac is non-vanishing at all points in O1 . Indeed, we have

∇a f = −2 χn ∇a χn = −2 χn δa n = −2 χa , and, therefore, T ac

=

∇c ∇a f − g ac (∇b ∇b f ) = −2 ∇c χa + 2 g ac (∇b χb )

=

−2 g ca + 2 g ac δb b = −2 g ac + 8 g ac = 6 g ac

throughout O1 .

Figure 2.5.2: A non-geodesic timelike curve enclosed in a tube (as considered in proposition 2.5.3).

One point about the proof deserves comment. As restricted to O1 and to M −cl(O), the field Tab that

we construct does satisfy the strengthened dominant energy condition. (In the first case, Tab = 6 gab , and in the second case, Tab = 0.) But we know — from the Geroch-Jang theorem itself — that it cannot satisfy that condition everywhere. So it must fail to do so in the buffer zone O−cl(O1 ). That shows us something. We can certainly choose f in the zone so that it smoothly joins with our choices for f on O1 and M −cl(O). But, no matter how clever we are, we cannot do so in such a way that T ab (as expressed

in (2.5.1)) satisfies the strengthened dominant energy condition.

Now we consider two examples of matter fields – perfect fluids in this section, and electromagnetic fields in the next. Perfect fluids are represented by three objects: a smooth four-velocity field η a , a smooth energy density field ρ, and a smooth isotropic pressure field p (the latter two as determined by a “co-moving” observer at rest in the fluid). In the special case where the pressure p vanishes, one speaks of a dust field. Particular instances of perfect fluids are characterized by “equations of state” that specify p as

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a function of ρ. (Specifically excluded here are such complicating factors as anisotropic pressure, shear stress, and viscosity.) Though ρ is generally assumed to be non-negative, some perfect fluids (e.g., to a good approximation, water) can exert negative pressure. The energy-momentum tensor field associated with a perfect fluid is Tab = ρ ηa ηb − p (gab − ηa ηb ).

(2.5.2)

So the energy-momentum density vector of the fluid at any point as determined by a co-moving observer (i.e., as determined relative to η a ) is T ab η b = ρ η a . In the case of a perfect fluid, the weak energy condition, dominant energy condition, and conservation condition come out as follows.16 WEC ⇐⇒ DEC

⇐⇒

CC

⇐⇒

ρ≥0

and p ≥ −ρ

|p| ≤ ρ   (ρ + p) η a ∇ η b − (g ab − η b η a ) ∇ p = a a  η a ∇a ρ + (ρ + p) (∇a η a ) =

0 0

First we verify the equivalences for the WEC and CC. (The one for the DEC is left as an exercise.) Then we make a few remarks about the physical interpretation of the two conditions jointly equivalent to CC. (WEC) Clearly, the WEC holds at a point q in M iff Tab ξ a ξ b ≥ 0 for all unit timelike vectors ξ a at

q. (If the inequality holds for all unit timelike vectors, it holds for all timelike vectors.) It is convenient to work with the condition in this form. If Tab is given by (2.5.2), and ξ a is a unit timelike vector at q, then Tab ξ a ξ b = (ρ + p)(η a ξa )2 − p. So

the WEC holds at q in M iff, for all such vectors ξ a at q,

(ρ + p)(η a ξa )2 − p ≥ 0.

(2.5.3)

Assume first that (ρ + p) ≥ 0 and ρ ≥ 0, and let ξ a be a unit timelike vector at q. Then, by the wrong-

way Schwarz inequality (proposition 2.2.3), (η a ξa )2 ≥ kη a k2 kξ a k2 = 1. Hence, (ρ + p)(η a ξa )2 − p ≥

(ρ+p)−p = ρ ≥ 0. So we have (2.5.3). Conversely, assume (2.5.3) holds for all unit timelike vectors ξ a at q. Then, in particular, it holds if ξ a = η a , and in this case we have 0 ≤ (ρ+p)(η a ηa )2 −p = (ρ+p)−p = ρ.

Note next that there is no upper bound to the value of (η a ξa )2 as ξ a ranges over unit timelike vectors

at q. (For example, let σ a be any unit spacelike vector at q orthogonal to η a , and let ξ a be of the form ξ a = (cosh θ) η a − (sinh θ) σ a , where θ is a real number. Then ξ a is a unit timelike vector, and

(η a ξa )2 = cosh2 θ. The latter goes to infinity as θ does.) So (2.5.3) cannot possibly hold for all unit timelike vectors at q unless (ρ + p) ≥ 0. This gives us the stated equivalence for the WEC. 16 The

dominant energy condition and the strengthened dominant energy condition are not equivalent in general, as we

have seen. But they are equivalent when applied, specifically, to perfect fluids. See problem 2.5.3.


(CC)

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If Tab is given by (2.5.2), then a straight forward computation shows that the conservation

condition (∇a T ab = 0) holds iff ρ (η a ∇a η b ) + ρ η b ∇a η a + η b (η a ∇a ρ) − (∇a p)(g ab − η a η b ) + p(η a ∇a η b ) + p η b ∇a η a = 0.

(2.5.4)

Assume that (2.5.4) does hold. Then contraction with ηb yields η a ∇a ρ + (ρ + p)(∇a η a ) = 0.

(2.5.5)

(Here we use the fact that the unit timelike vector field η b is orthogonal to its associated acceleration field η a ∇a η b and to its associated projection field hab = (gab − ηa ηb ).) And if we multiply (2.5.5) by η b

and then subtract the result from (2.5.4), we arrive at

(ρ + p) η a ∇a η b − (g ab − η b η a ) ∇a p = 0.

(2.5.6)

Thus (2.5.4) holds only if (2.5.5) and (2.5.6) do. And the converse is immediate. So we have our stated equivalence for the conservation condition. Problem 2.5.3. (i) Prove the stated equivalence for the DEC. (ii) Prove that, as restricted to perfect fluids, the SDEC is equivalent to the DEC. Now consider the physical interpretation of the two equations jointly equivalent to CC. (2.5.6) is the equation of motion for a perfect fluid. We can think of it as a relativistic version of Euler’s equation. (2.5.5) is an equation of continuity (or conservation) in the sense familiar from classical fluid mechanics. It is easiest to think about the special case of a dust field (p = 0). In this case, the equation of motion reduces to the geodesic equation η b ∇b η a = 0. That makes sense. In the absence of pressure, particles in the fluid are free particles. And the conservation equation reduces to η b ∇b ρ + ρ (∇b η b ) = 0. The

first term gives the instantaneous rate of change of the fluid’s energy density, as determined by a comoving observer. The term ∇b η b gives the instantaneous rate of change of its volume, per unit volume,

as determined by that observer. (We shall justify this claim in section 2.8.) In a more familiar notation, dρ ρ dV d(ρV ) the equation might be written + = 0 or, equivalently, = 0. (Here we use s for elapsed ds V ds ds proper time.) It asserts that (in the absence of pressure, as determined by a co-moving observer) the energy contained in an (infinitesimal) fluid blob remains constant, even as its volume changes. In the general case, the situation is more complex because the pressure in the fluid contributes to its energy (as determined relative to particular observers), and hence to what might be called its “effective mass density”. (If you compress a fluid blob, it gets heavier.) In this case, the WEC comes out as the requirement that (ρ + p) ≥ 0 in addition to ρ ≥ 0. The equation of motion can be expressed as (ρ + p) η b ∇b η a = hab ∇b p,

(2.5.7)

where hab is the projection field (g ab − η a η b ). This is an instance of the “second law of motion” (2.4.13)

as applied to an (infinitesimal) blob of fluid. On the left we have “effective mass density × acceleration”.


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On the right, we have the force acting on the blob, as determined by a co-moving observer. We can think of it as minus the gradient of the pressure (as determined by a co-moving observer). (The minus sign comes in because of our sign conventions.) Again, this makes sense. If the pressure on the left side of the blob is greater than that on the right, it will accelerate to the right. And in the general case we are now considering —where the pressure p need not vanish — the term (p ∇b η b ) in the conservation equation is required because the energy of the blob is not constant when its

volume changes as a result of the pressure. The equation governs the contribution made to its energy by pressure.

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2.6

Electromagnetic Fields

In this section we briefly discuss electromagnetic fields. Though our principal interest here is in the energy-momentum field Tab associated with them, we mention a few fundamental ideas of classical electromagnetic theory along the way. Electromagnetic fields are represented by smooth, anti-symmetric fields Fab (on the background spacetime (M, gab )). If a particle with mass m > 0, charge q, and four-velocity field ξ a is present, the force exerted by the field on the particle at a point is given by q F ab ξ b . (This condition uniquely characterizes Fab .) As noted at the end of section 2.4, if we use this expression for the force term in the relativistic version of “Newton’s second law” (2.4.13), we arrive at the Lorentz law of motion: q F ab ξ b = m ξ b ∇b ξ a .

(2.6.1)

It describes the motion of a charged particle in an electromagnetic field (at least when the contribution of the particle’s own charge to the field is negligible and may be ignored). Note again that the equation makes geometric sense. The acceleration vector on the right is orthogonal to ξ a . But so is the force vector on the left since Fab is anti-symmetric. The fundamental field equations of electromagnetic theory (“Maxwell’s equations”) are given by ∇[a Fbc] ∇a F

ab

= =

0

(2.6.2) b

J .

(2.6.3)

Here J a is the charge-current density field. It is characterized by the following condition: given any background observer at a point with four-velocity ξ a , J a ξa is the charge density there (arising from whatever charged matter is present) as determined by that observer. For example, in the case of a charged dust field, J a = µ η a , where η a is the four-velocity of the dust, and µ is its charge density as measured by a co-moving observer. Thus, if equation (2.6.1) expresses the action of the electromagnetic field on a charged (test) particle, equation (2.6.3) expresses the reciprocal action of charged matter on the field. The former acts a source for the latter. An important constraint on the charge-curent density field J a follows immediately from (2.6.3). Since F ab is anti-symmetric, ∇a J a = ∇a ∇n F na = ∇[a ∇n] F na . But 2 ∇[a ∇n] F na

= =

−F ma Rnman − F nm Raman −F mn Rmn + F mn Rnm

= =

−F ma Rma + F nm Rmn 0.

(The first two equalities follow, respectively, from clauses (1) and (2) of proposition 1.8.2. The third involves nothing more than a systematic change of abstract indices. The final equality follows from the symmetry of the Ricci tensor field.) So ∇a J a = 0.

(2.6.4)

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We can understand this as an assertion of the local conservation of charge. Notice that in the case of charged dust field with J a = µ η a , (2.6.4) comes out as η b ∇b µ + µ (∇b η b ) = 0.

(2.6.5)

This has exactly the same form as (2.5.5) in the special case where p = 0, and it can be analyzed in exactly the same manner. It asserts that, as determined by a co-moving observer, the total charge in an (infinitesimal) blob of charged dust remains constant, even as its volume changes. Problem 2.6.1. Show that Maxwell’s equations in the source free case (J a = 0) are conformally invariant, i.e., if an anti-symmetric field Fab satisfies them with respect to a metric gab , then it does so as ′ well with respect to any metric of the form gab = Ω2 gab . (Note: Here we need the fact that the dimen-

sion n of the background spacetime is 4. Hint: The conformal invariance of the first Maxwell equation (∇[a Fbc] = 0) follows immediately from problem 1.7.2 and does not depend on the value of n. To establish that of the second (∇a F ab = 0), use proposition 1.9.5 to show that ∇′ a (g ′am g ′bn Fmn ) =

1 (n − 4) ab (∇a F ab ) + F ∇a Ω, Ω4 Ω5

′ ′ where g ′ab = Ω−2 g ab is the inverse of gab , and ∇′ is the derivative operator compatible with gab .)

The energy-momentum tensor field associated with Fab is given by Tab = Fam F mb +

1 gab (Fmn F mn ). 4

(2.6.6)

We can gain some insight by introducing a reference observer O at a point p, with four-velocity ξ a , and considering the decomposition of Fab there into its “electric” and “magnetic” components. Let hab be the spatial projection tensor at the point determined by ξ a (defined by (2.4.3)). Further, let ǫabcd be a volume element on some open set containing p. Then we define µ =

J a ξa

(2.6.7)

ja

=

hab J b

(2.6.8)

Ea

=

(2.6.9)

Ba

=

ǫabc

=

F ab ξ b 1 abcd ǫ ξb Fcd 2 ǫabcn ξ n .

(2.6.10) (2.6.11)

E a and B a are, respectively, the electric and magnetic field vectors at the point as determined relative to O. (Clearly, if we had chosen the other volume element, −ǫabcd , we would have ended up with −B a . A

choice of volume element is tantamount to a choice of “right-hand rule”.) µ and j a are, respectively, the

charge density and current density vectors as determined relative to O. Note that E a , B a , and j a are all orthogonal to ξ a . We can think of ǫabc as a three-dimensional volume element defined on the orthogonal subspace of ξ a . (It is anti-symmetric, it is orthogonal to ξ a in all indices and, as one can show using (1.11.8) and it satisfies the normalization condition ǫabc ǫabc = −3!.)

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Reversing direction, we can recover Fab and J a from E a , B a , µ, and j a as follows: Ja

=

µ ξa + j a

(2.6.12)

Fab

=

2 E[a ξb] + ǫabcd ξ c B d .

(2.6.13)

The first assertion is an immediate consequence of the definitions of j a and µ. To verify the second, we substitute for B d on the right side. By (1.11.8), the anti-symmetry of Fab , and the definition of E a , we have 1 2 E[a ξb] + ǫabcd ξ c ( ǫdpqr ξp Fqr ) 2

1 (3!) ξ c δ [pa δ qb δ r]c ξp Fqr 2 + 3 ξ c ξ[a Fbc]

= 2 E[a ξb] + = 2 E[a ξb]

= 2 E[a ξb] + ξ c (ξa Fbc + ξc Fab − ξb Fac ) = Fab . Let us now return to our expression (2.6.6) for the energy-momentum field Tab . Our observer O with four-velocity ξ a will attribute to the electromagnetic field a four-momentum density T a b ξ b = F am Fmb ξ b +

1 a mn ξ (F Fmn ). 4

(2.6.14)

We can express the right side in terms of the relative electric and magnetic vectors E a and B a determined by O. (The computations are much like that used to prove (2.6.13).) We have F am Fmb ξ b

= F am Em = (2 E [a ξ m] + ǫampr ξp Br ) Em = −ξ a E m Em − ǫamr Em Br

(2.6.15)

and also F mn Fmn

=

(2 E [m ξ n] + ǫmnpq ξp Bq ) (2 E[m ξn] + ǫmnrs ξ r B s )

=

2 E n En + ǫmnrs ǫmnpq ξp Bq ξ r B s

=

2 E n En − 4 δ [pr δ q]s ξp Bq ξ r B s = 2 (E n En − B n Bn ).

(2.6.16)

Hence, T a b ξb =

1 (−E n En − B n Bn ) ξ a − ǫamr Em Br . 2

(2.6.17)

The coefficient of ξ a on the right side is the energy density of the field as determined by O. Using our notation for vector norms and temporarily dropping indices (and remembering that both E a and B a are spacelike (or the zero vector)), we can express it as

1 2

(kEk2 + kBk2 ). This will be familiar as the

standard textbook expression for the energy density of an electromagnetic field. The component of T a b ξ b

orthogonal to ξ a , namely −ǫamr Em Br , is the three-momentum density of the electromagnetic field as

determined by O. In more familiar vector notation (recall our discussion in section 1.11) it comes out as −(E × B). (E × B is called the “Poynting vector”.) Note that we can also work backward and derive (2.6.6), our expression for Tab , from the assumption

that (2.6.17) holds for all observers with four-velocity ξ a . (Reversing the calculation, one shows that

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(2.6.14) or, equivalently, (T a b − (F am Fmb +

1 4

g a b F mn Fmn )) ξ b = 0, holds for all unit timelike vectors

ξ a . (2.6.6) then follows by proposition 2.1.3.) So Tab is fully determined by the requirement that it code

values for

1 2

(kEk2 + kBk2 ) and −(E × B) for all observers.

Problem 2.6.2. Textbooks standardly assert that (kEk2 − kBk2 ) and E · B are relativistically invariant

(i.e., have common values for all observers). To verify this, it suffices to note that (in our notation): (−E a Ea + B a Ba ) = E a Ba

=

1 − F ab Fab 2 1 abcd ǫ Fab Fcd . 8

(2.6.18) (2.6.19)

We have proved the first assertion (equation (2.6.16)). Prove the second. Now we consider our two energy conditions and the conservation condition. Given any future-directed, unit timelike vector ξ a at a point, with corresponding electric and magnetic field vectors E a and B a , we have Tab ξ a ξ b

=

(Tab ξ b )(T ac ξc ) = ∇a T ab

=

1 (−E n En − B n Bn ) 2 1 n (E En − B n Bn )2 + (E n Bn )2 4 Ja F ab

(2.6.20) (2.6.21) (2.6.22)

The first follows immediately from (2.6.17) (and the fact that ǫabc is orthogonal to ξ a in all indices). We leave the second as an exercise. For the third, note that ∇a T ab

= = = =

1 ab g Fmn F mn ) 4 1 F am ∇a Fm b + Fm b ∇a F am + Fmn ∇b F mn 2 1 1 Fam (∇a F mb − ∇m F ab ) + Fm b ∇a F am + Fma ∇b F ma 2 2 1 a bm m ab b ma − Fam (∇ F + ∇ F + ∇ F ) + Fm b J m = Jm F mb . 2

∇a (F am Fm b +

(We get the third equality by systematically changing indices and using the anti-symmetry of Fab : Fam ∇a F mb = Fma ∇m F ab = −Fam ∇m F ab . We get the fourth and fifth from Maxwell’s equations

(first ∇a F am = J m , then ∇[a F bm] = 0) and, again, the anti-symmetry of Fab .)

Problem 2.6.3. Prove (2.6.21). (It follows immediately from this result that Tab ξ b is null iff E a Ea = B a Ba and E a Ba = 0. By problem 2.6.2, these conditions hold as determined relative to one unit timelike vector ξ a at a point iff they hold for all such vectors there. When they do hold (at all points), we say that Fab is a “null” field.) Maxwell’s equations play no role in the proof of (2.6.20) and (2.6.21). So we see that for any antisymmetric field Fab , the corresponding energy-momentum field Tab = Fam F mb + 41 gab (Fmn F mn ) satisfies both the weak and dominant energy conditions (since E a and B a are always spacelike or equal to the

137


zero vector 0). And it satisfies the strengthened dominant energy condition except in the special case where Fab is a non-vanishing null electromagnetic field (in the sense of problem 2.6.3). The situation is different with the conservation condition, for which Maxwell’s equations are essential. Suppose that the pair (Fab , J a ) satisfies them (and, therefore, that (2.6.22) holds). There are two cases to consider. If J a = 0, i.e., if no sources are present, then the conservation condition ∇a T ab = 0 is

automatically satisfied. But when charged matter is present, there is the possibility of energy-momentum being transferred from the electromagnetic field to that matter. So it should not be the energy-momentum of the electromagnetic field alone that is conserved. Instead, it should be the total energy-momentum present (arising from both field and charged matter) that is. By way of example, consider the case where a charged dust field serves as a source for the electromagnetic field. Suppose the dust is characterized by four-velocity field η a , mass density ρ, and charge density µ, the latter two as determined by a co-moving observer. Then we have J a = µ η a , and the energy-momentum field for the dust (alone) is given by ρ η a η b . So the total energy-momentum field in this case is given by Tab = Fam F mb +

1 gab (Fmn F mn ) + ρ ηa ηb . 4

(2.6.23)

Hence, by (2.6.22), ∇a T ab

=

Ja F ab + ∇a (ρ η a η b )

=

µ ηa F ab + ρ (η a ∇a η b ) + ρ η b ∇a η a + η b (η a ∇a ρ).

(2.6.24)

This is the counterpart to (2.5.4) that we considered in our discussion of perfect fluids. Arguing much as we did there, we can verify that in the present case we have the following equivalence. (Set the right hand side to 0, contract with ηb , and then subtract the resultant equation from the original.)   µ F b ηa = a CC ⇐⇒  η a ∇a ρ + ρ (∇a η a ) =

ρ (η a ∇a η b ) 0

The second equation on the right side is just (2.5.5) in the case where p = 0. It asserts that, as determined by a co-moving observer, the energy in an (infinitesimal) blob of dust remains constant, even as the volume of the blob changes. (Note that it also has exactly the same form as (2.6.5), which makes a corresponding assertion about charge conservation.) The first equation on the right side is an equation of motion for the dust field. It has exactly the same form as (2.6.1). It asserts, in a sense, that individual particles in the dust field obey the Lorentz law of motion. Thus, the energy-momentum of the electromagnetic field Fab fails to be conserved only to the extent it exerts a force on those particles and causes them to accelerate. As an afterthought, now, we recover the standard textbook formulation of Maxwell’s (four) equations from our formulation. To do so, we need a bit of structure in the background. Let us temporarily assume that (M, gab ) is not just any (temporally oriented) spacetime, but one that admits a future-directed, unit timelike vector field ξ a that is constant (∇a ξ b = 0). Let µ, j a , E a , B a , and ǫabc be as defined

138


above. Further, let D be the derivative operator induced on hypersurfaces orthogonal to ξ a . (Recall our discussion in section 1.10.) Then we have the following equivalences.   Db B b = 0 ∇[a Fbc] = 0 ⇐⇒  ǫabc Db Ec = −ξ b ∇b B a ∇a F ab = J b

  Db E b

⇐⇒

 ǫabc Db Bc

(∇ · B

=

0)

(∇ × E =

−

=

µ

(∇ · E

=

ξ b ∇b E a + j a .

(∇ × B

∂B ) ∂t

= µ) ∂E + j) = ∂t

(In each case, we have indicated how the right-side equation is formulated in standard (three-dimensional) vector notation.) We prove the first equivalence and leave the second as an exercise. Note first that by (2.6.13) and (1.11.8), we have ǫabcd Fcd

= ǫabcd (2 E[c ξd] + ǫcdrs ξ r B s ) = 2 ǫabcd Ec ξd − 4 δ [ar δ b] s ξ r B s = 2 ǫabcd Ec ξd − 2 ξ a B b + 2 ξ b B a .

Hence, since ξ a is constant, ǫabcd ∇b Fcd = ∇b (ǫabcd Fcd ) = 2 ǫabcd ξd ∇b Ec − 2 ξ a ∇b B b + 2 ξ b ∇b B a .

(2.6.25)

And for that same reason, ∇a hbc = ∇a (gbc − ξb ξc ) = 0. So, since ha b E b = E a and ha b B b = B a , Db B b ǫ

abc

Db Ec

= =

hmb hbn ∇m B n = ∇m (hmb hbn B n ) = ∇m B m = ∇b B b ǫ

abcd

ξd hmb

hnc ∇m En

=ǫ

abcd

ξd g mb ∇m (hnc En )

= ǫ

abcd

ξd ∇b Ec .

(2.6.26) (2.6.27)

(For the second equality in (2.6.27), note that ǫdabc ξd ξb = 0 and, hence, that ǫdabc ξd hmb = ǫdabc ξd g mb .) If we now replace ∇b B b and ǫabcd ξd ∇b Ec in (2.6.25) using (2.6.26) and (2.6.27), we arrive at ǫabcd ∇b Fcd = −2 ξ a (Db B b ) + 2 (ǫabc Db Ec + ξ b ∇b B a ).

(2.6.28)

Now ∇[a Fbc] = 0 iff ǫabcd ∇b Fcd = 0. (Why?) And the latter condition holds iff the sum on the right side of (2.6.28) is 0. But that sum consists of two terms, one tangent to ξ a and one orthogonal to ξ a . So

the sum is 0 iff both terms are 0. Thus we are left with the conclusion that ∇[a Fbc] = 0 iff Db B b = 0

and ǫabc Db Ec + ξ b ∇b B a = 0.

Problem 2.6.4. Prove the second equivalence (for ∇a F ab = J b ).

139


2.7

Einstein’s Equation

Once again, let (M, gab ) be our background relativistic spacetime with a specified temporal orientation. It is one of the fundamental ideas of relativity theory that spacetime structure is not a fixed backdrop against which the processes of physics unfold, but instead participates in that unfolding. It posits a dynamical interaction between the spacetime metric in any region and the matter fields there. The interaction is governed by Einstein’s field equation Rab −

1 R gab = 8 π Tab , 2

(2.7.1)

1 T gab ). 2

(2.7.2)

or, equivalently, Rab = 8 π (Tab −

Here Rab (= Rnabn ) is the Ricci tensor field, R (= Raa ) is the Riemann scalar curvature field, and T is the contracted field T aa .17 We start with four remarks about (2.7.1) and then consider two reformulations that provide a certain insight into the geometric significance of the equation. (1) It is sometimes taken to be a version of “Mach’s principle” that “the spacetime metric is uniquely determined by the distribution of matter”. And it is sometimes proposed that the principle can be captured in the requirement that “if one first specifies the energy-momentum distribution Tab on the spacetime manifold M , then there is exactly one (or at most one) Lorentzian metric gab on M that, together with Tab , satisfies (2.7.1)”. But there is a serious problem with the proposal. In general, one cannot specify the energy-momentum distribution in the absence of a spacetime metric. Indeed, in typical cases the metric enters explicitly in the expression for Tab . (Recall the expression (2.5.2) for a perfect fluid.) Thus, in looking for solutions to (2.7.1), one must, in general, solve simultaneously for the metric and matter field distribution. (2) Given any smooth metric gab on M , there certainly exists a smooth symmetric field Tab on M that, together with gab , is a solution to (2.7.1). It suffices to define Tab by the left side of the equation. But the field Tab so introduced will not, in general, be the energy-momentum field associated with any known matter field. And it will not, in general, satisfy the weak energy condition discussed in section 2.5. If the latter condition is imposed as a constraint on Tab , Einstein’s equation is an entirely non-trivial restriction on spacetime structure. Discussions of spacetime structure in classical relativity theory proceed on three levels according to the stringency of the constraints imposed on Tab . At the first level, one considers only “exact solutions”, i.e., solutions where Tab is, in fact, the aggregate energy-momentum field associated with one or more known matter fields. So, for example, one might undertake to find all perfect fluid solutions exhibiting particular symmetries. At the second level, one considers the larger class of what might be called “generic solutions”, 17 We

use “geometrical units” in which the gravitational constant G as well as the speed of light c are 1.


140

i.e., solutions where Tab satisfies one or more generic constraints (of which the weak and dominant energy conditions are examples). It is at this level, for example, that the singularity theorems of Penrose and Hawking (Hawking and Ellis [30]) are proved. Finally, at the third level, one drops all restrictions on Tab , and Einstein’s equation plays no role. Many results about global structure are proved at this level, e.g., the assertion that closed timelike curves exist in any relativistic spacetime (M, gab ) where M is compact. (3) We have presented Einstein’s equation in its original form. He famously added a “cosmological constant” term (−Λgab ) in 1917 to allow for the possibility of a static cosmological model with a perfect fluid source with p = 0 and ρ > 0.18 (We shall see why the addition is necessary under those conditions at the end of section 2.11.) But Einstein was never happy with the revised equation Rab − or, equivalently, Rab

1 R gab − Λ gab = 8 π Tab , 2

1 = 8 π Tab − T gab − Λ gab , 2

(2.7.3)

(2.7.4)

and was quick to revert to the original version after Hubble’s redshift observations gave convincing evidence that the universe is, in fact, expanding. After that, he thought, there was no need to have a static cosmological model. (That the theory suggested the possibility of cosmic expansion before Hubble’s observations must count as one of its great successes.) Since then the constant has often been reintroduced to help resolve discrepancies between theoretical prediction and observation, and then abandoned when the (apparent) discrepancies were resolved. (See Earman [14] for a masterful review of the history.) The story continues. Recent observations indicating an accelerating rate of cosmic expansion seem to imply that our universe is characterized by a positive value for Λ or something that mimics its effect. In what follows, we shall continue to write Einstein’s equation in the form (2.7.1) and think of the cosmological term as absorbed into the expression for the energy-momentum field Tab . The magnitude and physical interpretation of this contribution to Tab are topics of great importance in current physics.19 But they will play no role in our discussion. Problem 2.7.1. Equations (2.7.3) and (2.7.4) are equivalent only if the dimension n of the background manifold is 4. Show that in the general case (at least if n ≥ 3), inversion of (2.7.3) leads to 1 2 Rab = 8 π Tab − T gab − Λ gab . (n − 2) (n − 2)

(2.7.5)

(4) It is instructive to consider the relation of Einstein’s equation to Poisson’s equation ∇2 φ = 4 π ρ,

(2.7.6)

the field equation of Newtonian gravitation theory. Here φ is the Newtonian gravitational potential, and ρ is the Newtonian mass density function. In the geometrized formulation of the theory that we shall 18 He 19 See

did so for other reasons as well (see Earman [14]), but we pass over them here. Earman [14], once again, and references cited there.

141


consider in chapter 4, one trades in the potential φ in favor of a curved derivative operator and one recovers ρ from a mass-momentum field T ab . In the end, Poisson’s equation comes out as Rab = 8 π (Tâb −

1 tab Tˆ). 2

(2.7.7)

Here Rab is the Ricci tensor field associated with the new curved derivative operator, tab is the temporal metric, Tâb = T mn tma tnb , and Tˆ = T mn tmn . (See (4.2.10) and the discussion that precedes it.) The resemblance to (2.7.2) is, of course, striking. It is particularly close in the special case where ρ = 0. For in this case, T ab = 0 and (2.7.7) reduces to Rab = 0. The latter is exactly the same as Einstein’s equation (2.7.2) in the empty space case. The geometrized formulation of Newtonian gravitation was discovered after general relativity in the 1920s. But now, after the fact, we can put ourselves in the position of a hypothetical investigator who is considering possible candidates for a relativistic field equation and who knows about the geometrized formulation of Newtonian theory. What could be more natural than to adapt (2.7.7) and simply replace tab with gab ? This seems to me one of the nicest routes to Einstein’s equation (2.7.2). Again, the route is particularly direct in the empty space case. For then one starts with the Newtonian empty space equation (Rab = 0) and simply leaves it intact. Let us now put aside the question of how one might try to motivate Einstein’s equation, and consider two reformulations. Let ξ a be a unit timelike vector at a point p in M , and let S be a spacelike hypersurface containing p that is orthogonal to ξ a there. (We understand a hypersurface in M to be spacelike if, at every point, vectors tangent to the surface are spacelike. This condition guarantees that the hypersurface is metric. (Recall our discussion in section 1.10.)) Further, let hab and πab be the first and second fundamental forms on S, and let D be the derivative operator on S determined by hab . Associated with D is a Riemann curvature field Rabcd on S. We know (recall (1.10.21)) that the contracted scalar field R = Rabca hbc

satisfies

R = π 2 − πab π ab + R − 2 Rnr ξ n ξ r

(2.7.8)

at p. In the special case where S has vanishing extrinsic curvature (πab = 0) at p, this can be expressed as

1 1 gab R) ξ a ξ b = − R. 2 2 If Einstein’s equation holds, it therefore follows that (Rab −

R = −16π (Tab ξ a ξ b ).20 20 There

(2.7.9)

(2.7.10)

is an issue here of sign convention that is potentially confusing. We seem to be led to the conclusion that the

Riemann scalar curvature of S is less than or equal to 0 — at least if Tab satisfies the weak energy condition. But it might be more natural to say that it is greater than or equal to 0. We are working here with R as determined relative to the

negative definite metric hab , and a sign flip is introduced if we work, instead, with the positive definite metric −hab . The

switch from hab to −hab leaves D, Ra bcd , and Rcd intact but reverses the sign of R = hbc Rbc .

142


One can also work backwards. Suppose (2.7.10) holds for all unit timelike vectors at p and all orthogonal spacelike hypersurfaces through p with vanishing extrinsic curvature there. Then, by (2.7.9), it must be the case that (Rab −

1 R gab ) ξ a ξ b = 8 π Tab ξ a ξ b 2

(2.7.11)

for all unit timelike vectors ξ a at p. This, in turn, implies Einstein’s equation (by lemma 2.1.3). So we have the following equivalence. (⋆) Einstein’s equation Rab − 12 R gab = 8 π Tab holds at p iff for all unit timelike vectors ξ a

at p, and all orthogonal spacelike hypersurfaces S through p with vanishing extrinsic curvature

there, the scalar curvature of S at p is given by R = −16π (Tab ξ a ξ b ). We can give the result a somewhat more concrete formulation by casting it in terms of a particular class of spacelike hypersurfaces. Consider the set of all geodesics through p that are orthogonal to ξ a there. The (images of these) curves, at least when restricted to a sufficiently small open set containing p, sweep out a smooth spacelike hypersurface that is orthogonal to ξ a at p.21 (See figure 2.7.1.) We

Figure 2.7.1: A “geodesic generated hypersurface” through a point is constructed by projecting geodesics in all directions orthogonal to a given timelike vector there.

shall call it a geodesic generated hypersurface. (We cannot speak of the geodesic generated hypersurface through p orthogonal to ξ a because we have left open how far the generating geodesics are extended. But given any two, their restrictions to a suitably small open set containing p coincide.) Geodesic generated hypersurfaces are of interest in their own right, the present context aside, because they are natural candidates for a notion of “local simultaneity slice” (as determined relative to a timelike vector at a point). We can think of them as instances of private space. (The contrast here is with public space which is determined not relative to a single timelike vector or timelike curve, but relative to a congruence of timelike curves. For more on this difference between private space and public space, see Rindler [53, 54] and Page [49].) Now suppose S is a geodesic generated hypersurface generated from p. We claim that it has vanishing extrinsic curvature there. We can verify this with a simple calculation very much like that used to prove 21 More

precisely, let Sp be the spacelike hyperplane in Mp orthogonal to ξ a . Then for any sufficiently small open set O

in Mp containing p, the image of (Sp ∩ O) under the exponential map exp : O → M is a smooth spacelike hypersurface in

M containing p that is orthogonal to ξ a there. (See, for example, Hawking and Ellis [30, p. 33].)

143


proposition 1.10.7. Let ξ a be a smooth, future-directed, unit timelike field, defined on some open subset of S containing p, that is orthogonal to S. Let hab be the corresponding projection field on S. Further, let σ a be the tangent field to a geodesic (relative to ∇) through p that is orthogonal there to ξ a . Then

along the image of the geodesic we have σ a ∇a σ b = 0 and σ a ξa = 0 (or, equivalently, ha b σ b = σ a ). The

latter holds because the image of the geodesic is contained in S and so is everywhere orthogonal to its normal field. Hence, by (1.10.16), we have πab σ a σ b

= (hm a hn b ∇m ξn ) σ a σ b = σ m σ n ∇m ξn = σ m ∇m (σ n ξn ) − ξn σ m ∇m σ n = 0

along the image of the geodesic. In particular, the condition holds at p. But given any vector at p orthogonal to ξ a , we can choose our initial geodesic so that it has that vector for its tangent at p. Hence, πab σ a σ b = 0 at p for all such orthogonal vectors. Since πab is symmetric, as well as orthogonal to the normal field ξ a , it follows that πab = 0 at p. Consider again the equivalence (⋆). If we rerun the argument used before, but systematically cast it in terms of geodesic generated hypersurfaces, we arrive at the following alternate formulation. Proposition 2.7.1. Let Tab be a smooth symmetric field on M , and let p be a point in M . Then Einstein’s equation Rab −

1 2

R gab = 8 π Tab holds at p iff for all unit timelike vectors ξ a at p, and all

geodesic hypersurfaces S generated from p that are orthogonal to ξ a , the scalar curvature of S at p is

given by R = −16π (Tab ξ a ξ b ). Our second reformulation of Einstein’s equation is phrased in terms of geodesic deviation. Let ξ a be a smooth, future-directed, unit timelike vector field whose associated integral curves are geodesics, i.e., a geodesic reference frame. Further, let λa be a vector field on one of the integral curves γ satisfying £ξ λa = 0. (So ξ b ∇b λa = λb ∇b ξa .) Finally, assume λa is orthogonal to ξ a at some point on γ. Then

it must be orthogonal to the latter at all points on γ. This follows because the inner product (ξ a λa ) is constant on γ: ξ b ∇b (ξ a λa ) = λa ξ b ∇b ξ a + ξ a ξ b ∇b λa = ξ a ξ b ∇b λa = ξ a λb ∇b ξa =

1 b 1 λ ∇b (ξ a ξa ) = λb ∇b (1) = 0. 2 2

We can think of λa as a connecting field that joins the image of γ to the image of another, “infinitesimally close”, integral curve of ξ a . Then the field ξ n ∇n (ξ m ∇m λa ) represents the acceleration of the latter relative to γ. We know from proposition 1.8.5 that it satisfies the “equation of geodesic deviation”: ξ n ∇n (ξ m ∇m λa ) = Ra bcd ξ b λc ξ d .

(2.7.12) i

Now we define the “average radial acceleration” of ξ a at a point p on γ. Let λa (i = 1, 2, 3) be any three 1

2

3

connecting fields (as just described) such that, at p, the vectors ξ a , λa , λa , λa form an orthonormal set.

144


i

For each i, the (outward-directed) radial component of the relative acceleration vector ξ n ∇n (ξ m ∇m λa ), i

i.e., its component in the direction λ, has magnitude

i

i

− λa ξ n ∇n (ξ m ∇m λa ). (We need the minus sign because λa is spacelike.) We now take the average radial acceleration (ARA) of ξ a at p to be ARA = −

3 i 1 X i a n m λa ξ ∇n (ξ ∇m λ ). 3 i=1

(2.7.13)

Of course, we need to check that the sum on the right side is independent of our initial choice of i i P connecting fields. The orthonormality condition implies that at p we have gac = ξa ξc − 3i=1 λa λc .

Hence, by (2.7.12), we also have ARA = −

3 3 X i i i 1 a 1 a 1 X i c a b c d b d b d c c λa λ ) = − R bcd ξ ξ (ξa ξ − ga ) λa R bcd ξ λ ξ = − R bcd ξ ξ ( 3 i=1 3 3 i=1

at p. But Rabcd ξ c ξ d = 0, and Rabcd gac = Rabad = −Rabda = −Rbd . So we may conclude that 1 ARA = − Rbd ξ b ξ d 3

(2.7.14)

holds at p. Thus, as claimed, average radial acceleration is well defined. Now if Einstein’s equation holds at p, it follows that ARA = −

1 8π (Tab − T gab )ξ a ξ b 3 2

(2.7.15)

holds there as well. And conversely, if (2.7.15) holds at p for all geodesic reference frames, then it must be the case, by (2.7.14), that Rbd ξ b ξ d = 8 π (Tab − 12 T gab )ξ b ξ d holds for all unit timelike vectors ξ a there. And this, in turn, implies that Einstein’s equation holds at p. So we have the following equivalence.

Proposition 2.7.2. Let Tab be a smooth symmetric field on M , and let p be a point in M . Then Einstein’s equation Rab − 21 R gab = 8 π Tab holds at p iff for all geodesic reference frames ξ a (defined on some open 1 8π (Tab − T gab )ξ a ξ b . set containing p), the average radial acceleration of ξ a at p is given by ARA = − 3 2 We considered three energy conditions (weak, dominant, and strengthened dominant) in section 2.5. Let us now consider a fourth. Let Tab be the energy-momentum field associated with a matter field F . Strong Energy Condition (SEC): Given any timelike vector ξ a at any point in M , 1 (Tab − T gab ) ξ a ξ b ≥ 0. 2 Equation (2.7.15) provides an interpretation. Suppose that Einstein’s equation holds. Then F satisfies

the strong energy condition iff, for all geodesic reference frames, the average (outward directed) radial acceleration of the frame is negative or 0. This captures the claim, in a sense, that the “gravitational field” generated by F is “attractive”.


145

Problem 2.7.2. Give examples of each of the following. (1) A smooth symmetric field Tab that satisfies the SDEC (and so also the WEC and DEC) but not the SEC (2) A smooth symmetric field Tab that satisfies the SEC but not the WEC (and so not the DEC or SDEC either) Problem 2.7.3. Consider a perfect fluid with four-velocity η a , energy density ρ, and pressure p. Show that it satisfies the strong energy condition iff (ρ + p) ≥ 0 and (ρ + 3p) ≥ 0.


2.8

146

Fluid Flow

In this section, we consider fluid flow and develop the standard formalism for representing the rotation and expansion of a fluid at a point. (Later, in sections 3.2 and 3.3, we shall consider several different notions of global rotation.) Once again, let (M, gab ) be our background relativistic spacetime. We are assuming it is temporally orientable and endowed with a particular temporal orientation. Let ξ a be a smooth, future-directed unit timelike vector field on M (or some open subset of M ). We understand it to represent the four-velocity field of a fluid. Further, let hab be the spatial projection field determined by ξ a . The rotation and expansion fields associated with ξ a are defined as follows: ωab

= h[am hb] n ∇m ξn

(2.8.1)

θab

= h(am hb)n ∇m ξn .

(2.8.2)

They are smooth fields, orthogonal to ξ a in both indices, and satisfy ∇a ξb = ωab + θab + ξa (ξ m ∇m ξb ).

(2.8.3)

(This follows since ωab + θab = ha m hb n ∇m ξn = (ga m − ξa ξ m ) (gb n − ξb ξ n ) ∇m ξn , and ξ n ∇m ξn = 0.) Our first task is to give the two fields a geometric interpretation and, in so doing, justify our terminology. We start with the rotation field ωab .

Let γ be an integral curve of ξ a , and let p be a point on the image of γ. Further, let η a be a vector field on the image of γ that is “carried along by the flow of ξ a ” (i.e., £ξ η a = 0) and is orthogonal to ξ a at p. (It need not be orthogonal to ξ a elsewhere.) We think of the image of γ as the worldline of a fluid element O, and think of η a at p as a “connecting vector” that spans the distance between O and a neighboring fluid element N that is “infinitesimally close”. The instantaneous velocity of N relative to O at p is given by ξ a ∇a η b . But ξ a ∇a η b = η a ∇a ξ b (since £ξ η a = 0). So, by (2.8.3), and the orthogonality

of ξ a with η a at p, we have

ξ a ∇a η b = (ωa b + θa b ) η a .

(2.8.4)

at the point. Here we have simply decomposed the relative velocity vector into two components. The first, (ωa b η a ), is orthogonal to η a since ωab is anti-symmetric. (See figure 2.8.1.) It is naturally understood as the instantaneous rotational velocity of N with respect to O at p. In support of this interpretation, consider the instantaneous rate of change of the squared length (−η b ηb ) of η a at p. It follows from (2.8.4) that ξ a ∇a (−η b ηb ) = − 2 θab η a η b .

(2.8.5)

147


Thus the rate of change depends solely on θab . Suppose θab = 0. Then the instantaneous velocity of N with respect to O at p has a vanishing radial component. If ωab 6= 0, N can still have non-zero velocity

there with respect to O. But it can only be a rotational velocity. The two conditions (θab = 0 and ωab 6= 0) jointly characterize “rigid rotation”.

The rotation tensor ωab at a point p determines both an (instantaneous) axis of rotation there, and an (instantaneous) speed of rotation. As we shall see, both pieces of information are built into the angular velocity (or twist ) vector ωa =

1 abcd ǫ ξb ωcd 2

(2.8.6)

at p. (Here ǫabcd is a volume element defined on some open set containing p. Clearly, if we switched from the volume element ǫabcd to its negation, the result would be to replace ω a with −ω a .) ωa

ωb a η b (rotational velocity vector) ρa

η a (connecting vector)

Figure 2.8.1: The angular velocity (or twist) vector ω a . It points in the direction of the instantaneous axis of rotation of the fluid. Its magnitude kω a k is the instantaneous angular speed of the fluid about that axis. Here η a connects the fluid element O to the “infinitesimally close” fluid element N . The rotational velocity of N relative to O is given by ωb a η b . The latter is orthogonal to η a .

If follows from (2.8.6) (and the anti-symmetry of ǫabcd ) that ω a is orthogonal to ξ a . It further follows that ωa ωab

1 abcd ǫ ξb ∇c ξd , 2 = ǫabcd ξ c ω d . =

(2.8.7) (2.8.8)

Hence, ωab = 0 iff ω a = 0. Both (2.8.7) and (2.8.8) are verified with simple calculations. We do the first and leave the second as an exercse. For the first, we have 2 ωa

=

ǫabcd ξb ωcd = ǫabcd ξb h[c r hd]s ∇r ξs = ǫabcd ξb hcr hds ∇r ξs

=

ǫabcd ξb gcr gds ∇r ξs = ǫabcd ξb ∇c ξd .


148

(The second equality follows from the anti-symmetry of ǫabcd , and the third from the fact that ǫabcd ξb is orthogonal to ξ a in all indices.) Notice that (2.8.6) has exactly the same form as our definition (2.6.10) of the magnetic field vector B a determined relative to a Maxwell field Fab and four-velocity vector ξ a (B a =

1 abcd ξb 2 ǫ

Fcd ). It is for this reason that the magnetic field is sometimes described as the “rotational

component of the electromagnetic field”. Problem 2.8.1. Prove (2.8.8) Problem 2.8.2. We have seen that the conditions (i) ωab = 0 and (ii) ω a = 0 are equivalent at any point. Show that they are also equivalent (at any point) with (iii) ξ[a ∇b ξc] = 0. We claim now that ω a points in the direction of the instantaneous axis of rotation (of the fluid flow associated with ξ a ). (See figure 2.8.1 again.) More precisely, with the connecting field η a as above, we show that, at p, ωb a η b = 0

⇐⇒

η a is proportional to ω a .

(2.8.9)

(Or, in the language of “infinitesimally close” fluid elements, the rotational velocity of N with respect to O vanishes iff the connecting vector from O to N is aligned with ω a .) The implication from right to left follows immediately from (2.8.8) (and the anti-symmetry of ǫabcd ). Conversely, suppose ωb a η b = 0. Then, by (2.8.8), 0

=

(ξn ωp ǫamnp ) ωba η b = ξn ωp ǫamnp ǫbacd ξ c ω d η b

=

3! δ

=

(η m ω p ωp − ω m η p ωp ).

[m n p] b bδ cδ dη

ξ c ω d ξn ωp = 3! η [m ξ n ω p] ξn ωp

(For the final equality, here we use the fact that ξ a is orthogonal to η a at p, and orthogonal to ω a everywhere.) Now if ω p ωp = 0, then ω a = 0. (The twist vector ω a is orthogonal to ξ a and, by proposition 2.2.1, the only null vector orthogonal to a timelike vector is the zero vector.) And in this case, η a is trivially aligned with ω a . So we may assume that ω p ωp 6= 0. It then follows that η a = k ω a , where

k = (ω p ηp )/(ω n ωn ).

Next, we claim that the magnitude of ω a is the instantaneous angular speed (of the fluid flow associated with ξ a ). The angular speed for the connecting vector η a is given by the ratio of the linear speed of rotation η b ωb a (i.e., the magnitude of ωb a η b ) to the magnitude of the radius vector ρa = η a − n ω . (See figure ω ωn 2.8.1 again.) (If ω n ωn = 0, then ωab = 0, and the speed of angular rotation is 0.) It follows with a bit of calculation much like that done previously in this section that (angular speed)2 = i.e., the angular speed is kω a k, as claimed.

−ωb a η b ωca η c = ... = (−ω n ωn ), −ρn ρn

(2.8.10)

Problem 2.8.3. Complete the calculation in (2.8.10). (Hint: Do not forget that we are doing the calculation at the initial point p where the connecting vector η a is orthogonal to ξ a .)

149


The two italicized conditions concerning, respectively, the orientation and magnitude of ω a , determine it up to sign. With the preceding remarks as motivation, we now say that our future-directed, unit timelike vector field ξ a is irrotational or twist-free at a point if ωab = 0 there (or, equivalently, if ω a = 0 or if ξ[a ∇b ξc] = 0 there). It will be instructive to consider a condition that captures the requirement that

ξ a is twist-free everywhere. Let us say that a timelike vector field ξ a (not necessarily of unit length)

is hypersurface orthogonal if there exist smooth, real valued maps f and g (with the same domains of definition as ξ a ) such that, at all points, ξa = f ∇a g. Note that if the condition is satisfied, then the

hypersurfaces of constant g value are everywhere orthogonal to ξ a . (For if σ a is a vector tangent to one of these hypersurfaces, σ n ∇n g = 0. So σ n ξn = σ n (f ∇n g) = 0.) Let us further say that ξ a is

locally hypersurface orthogonal if the restriction of ξ a to every sufficiently small open set is hypersurface orthogonal. Proposition 2.8.1. Let ξ a be a smooth, future-directed unit timelike vector field defined on M (or some open subset of M ). Then the following conditions are equivalent. (1) ωab = 0 everywhere. (2) ξ a is locally hypersurface orthogonal. Proof. The implication from (2) to (1) is immediate. For if ξa = f ∇a g, then ωab

=

h[am hb] n ∇m ξn = h[am hb] n ∇m (f ∇n g)

=

f h[am hb] n ∇m ∇n g + h[am hb] n (∇m f ) (∇n g)

=

f ha m hb

n

∇[m ∇n] g + ha m hb

n

(∇[m f ) (∇n] g).

But ∇[m ∇n] g = 0, since ∇ is torsion-free; and the second term in the final line vanishes as well since hb

n

∇n g = f −1 hb

n

ξn = 0. So ωab = 0. But the converse is non-trivial. It is a special case of Frobenius’

theorem (Wald [60, p. 436]). There is a nice picture that goes with the proposition. Think about an ordinary rope. In its natural twisted state, the rope cannot be sliced in such a way that the slice is orthogonal to all individual fibers. But if the rope is first untwisted, then such a slicing is possible. Thus orthogonal sliceability is equivalent to fiber-untwistedness. The proposition extends this intuitive equivalence to the four-dimensional “spacetime ropes” (i.e., congruences of worldlines) encountered in relativity theory. It asserts that a congruence is twist-free iff it is, at least locally, hypersurface orthogonal. Let us now switch our attention to the expansion tensor θab associated with ξ a . First, we decompose it into two pieces. We set θ

=

σab

=

θaa = ∇a ξ a 1 θab − hab θ, 3

(2.8.11) (2.8.12)

150


so that (2.8.3) can be expressed in the expanded form 1 hab θ + ξa (ξ n ∇n ξb ). 3

∇a ξb = ωab + σab +

(2.8.13)

Notice that the two expressions for θ in (2.8.11) are equal since ξ n ∇m ξn = 0 and, therefore θaa = g ab θab = g ab h(am hb)n ∇m ξn = hmn ∇m ξn = (g mn − ξ m ξ n ) ∇m ξn = ∇n ξ n . Notice too that σaa = 0, since σaa = θaa −

1 3

(2.8.14)

(gaa − ξa ξ a ) θ = θ − θ = 0. We call θ the scalar expansion field associated with ξ a

and call σab the shear tensor field associated with it. We can motivate this terminology much as we did that for ωab . We claim first that θ is a measure of the rate at which the volume of an (infinitesimal) blob of fluid increases under the flow associated with ξ a . (It is the counterpart to the “divergence” of a vector field in ordinary three-dimensional Euclidean vector analysis.) To justify this interpretation, we do a simple calculation. 1

2

3

Let γ be an integral curve of ξ a , and let p be any point on its image. Further, let η a , η a , η a be three i

vector fields on the image of γ that (i) are carried along by the flow associated with ξ a (i.e., £ξ η a = 0, for 1

1

2

2

3

3

i = 1, 2, 3), and (ii) together with ξ a , form an orthonormal basis at p. Then hab = −(η a η b + η a η b + η a η b )

at p. We consider the rate of change of the volume function V = ǫabcd ξ a η b η c η d in the direction ξ a . It 1

2

3

turns out that, at p, ξ n ∇n V = θ V.

(2.8.15)

It is in this sense that θ gives the instantaneous rate of volume increase, per unit volume, under the flow associated with ξ a . (This is the claim we made at the end of section 2.5.) i

i

i

To verify (2.8.15), we compute ξ n ∇n V . Since £ξ η a = 0, we have ξ n ∇n η a = η n ∇n ξ a and, hence, ξ n ∇n V

1

= ξ n ∇n (ǫabcd ξ a η b η c η d ) i h 1 2 3 3 1 2 = ǫabcd (ξ n ∇n ξ a ) η b η c η d + ... + (η n ∇n ξ d ) ξ a η b η c . 1

2

3

2

3

1

2

(2.8.16)

3

Now the vector ǫabcd η b η c η d is orthogonal to η b , η c , and η d . So, at p, it must be co-aligned with 1

2

3

1

2

3

ξa . Indeed, we have ǫabcd η b η c η d = (ǫnbcd ξ n η b η c η d ) ξa = V ξa ξa (ξ n ∇n ξ a )V = 0 at p. Similarly, for example, we have

ǫabcd (η n ∇n ξ b ) ξ a η c η d = −ηb (η n ∇n ξ b )V 2

1

3

1

1

at p. So, after handling all terms on the right side of (2.8.16) this way, we are left, at p, with ξ n ∇n V

=

h i 1 1 2 2 3 3 −V η r (η n ∇n ξ r ) + η r (η n ∇n ξ r ) + η r (η n ∇n ξ r )

=

−V (η r η n + η r η n + η r η n )(∇n ξ r ) = V hrn ∇n ξ r

=

V (grn − ξr ξ n ) ∇n ξ r = V ∇n ξ n = V θ.

1

1

2

2

3

1

2

3

there. So, ǫabcd (ξ n ∇n ξ a ) η b η c η d =

3

151


This gives us (2.8.15). Now consider σab . It is symmetric (and orthogonal to ξ a ). So we can choose our three vector fields 1

2

3

η a , η a , η a so that, in addition to being carried along by the flow of ξ a , and (with ξ a ) forming an or1

1

2

1

3

2

2

3

3

thonormal basis at p, they satisfy σab = −(k η a η b + k η a η b + k η a η b ) at p. (It is a basic fact of linear

algebra that we can find an orthonormal basis at p that diagonalizes the symmetric 4 × 4 matrix of i

i

i

i

i

σab -components.) Then σ ab η b = k η a , for each i, i.e., η a is an eigenvector of σ ab with eigenvalue k. And 1

i

1

1

2

2

2

3

3

1

3

2

3

the coefficients k sum to 0, since 0 = σaa = −(k η a η a + k η a η a + k η a η a ) = (k + k + k). i

Suppose for the moment that ωab = 0 and θ = 0 at p. Then, by (2.8.4) and (2.8.13), ξ n ∇n η a = i

i

i

i

σna η n = k η a , for all i, at p. So, if we think of η a as a “connecting vector” pointing from an observer O to an (infinitesimally) close neighbor N , then the instaneous velocity of N relative to O is directed i

1

2

3

radially away from O at p and has magnitude k there. Thus, each of the vectors η a , η a , η a is an axis i

of instantaneous expansion (or contraction) with associated magnitude k. Since the magnitudes sum to 0, expansion along one axis can occur only if there is contraction along another. Individual expansions and contractions so compensate each other that there is no net increase in volume. (Again, we are now considering the case where θ is 0.) i

In general, the expansion factors k are all different. But, for purposes of illustration, suppose that the 1

2

factors on two axes are equal — say k = k. Further imagine that our infinitesimal blob has the shape of a sphere at p. Then there are two possibilities. If the common factor is positive, then the action of the 3

flow flattens it into a pancake with axis η a (“pancake shear”). If it is negative, then it is elongated into a 3

hot dog with axis η a (“hot dog shear”). The second possibility is illustrated in figure 2.8.2, where three possible actions are illustrated.

ωab 6= 0 θ=0 σab = 0

ωab = 0 θ>0 σab = 0

ωab = 0 θ=0 σab 6= 0

ωa

rigid rotation uniform spherical expansion

volume-preserving shear

Figure 2.8.2: Rotation, Expansion, Shear

152


The full expansion tensor field θab can be given another interesting geometric interpretation in the case where it is associated with a unit timelike flow ξ a that is everywhere twist-free. In this case, by proposition 2.8.1, ξ a is, at least locally, hypersurface orthogonal. Let S be a spacelike hypersurface to which ξ a is orthogonal. The extrinsic curvature of S is given by πab = ham hbn ∇m ξn . (Recall (1.10.16).) But ham hbn ∇m ξn = ωab + θab , by (2.8.1) and (2.8.2). So in the present case (ωab = 0), we have

πab = θab . Thus, the expansion tensor field associated with a twist-free unit timelike field ξ a is just the extrinsic curvature of the spacelike hypersurfaces to which ξ a is orthogonal. This gives us another way to think about the extrinsic curvature of spacelike hypersurfaces. When

πab = 0, normal vectors to the surface do not recede from one another. “Connecting vectors” between “infinitesimally” close surface normals do not expand. (See figure 2.8.3.) But when πab 6= 0, connecting vectors do expand.

θab 6= 0

θab = 0

Figure 2.8.3: Expansion and Extrinsic Curvature

Finally, we derive an expression for the rate of change of the scalar expansion function θ (“Raychaudhuri’s equation”): ξ a ∇a θ = −Rab ξ a ξ b + ωab ω ab −

1 2 θ − σab σ ab + ∇a (ξ n ∇n ξ a ). 3

(2.8.17)

We shall need it later in section 2.11. (Here ξ a is still a smooth future-directed unit timelike vector field on our background spacetime (M, gab ).) The derivation proceeds in two steps. First, it follows from (1.8.1) that ξ a ∇a θ

=

ξ a ∇a ∇b ξ b = −ξ a Rbcab ξ c + ξ a ∇b ∇a ξ b

=

−Rca ξ c ξ a + ∇b (ξ a ∇a ξ b ) − (∇b ξ a )(∇a ξ b ).

Next, we evaluate the term (∇b ξ a )(∇a ξ b ) using the expansion in (2.8.13): ∇a ξb = ωab + σab +

1 3

hab θ +

ξa (ξ n ∇n ξb ). A straightforward computation establishes that (∇b ξa )(∇a ξ b ) = −ωab ω ab +

1 2 θ + σab σ ab . 3

(All terms involving ξa or ξb are 0 because hab , ωab , σab , and ξ n ∇n ξa are all orthogonal to ξ a in all indices.

The terms involving ωab together with either hab or σ ab are 0 because the former is anti-symmetric whereas the latter are symmetric. The terms involving hab and σab are 0 because σa a = 0.) This gives us (2.8.17).


2.9

153

Killing Fields and Conserved Quantities

In relativity theory, there is a natural association between Killing fields and conserved quantities. We consider it briefly in this section. Let κa be a smooth field on our background spacetime (M, gab ). Recall (section 1.9) that κa is said to be a Killing field if its associated local flow maps Γs are all isometries or, equivalently, if £κ gab = 0. The latter condition can also be expressed as ∇(a κb) = 0. Any number of standard symmetry conditions — local versions of them, at least22 — can be cast as claims about the existence of Killing fields. Here are a few examples. (M, gab ) is stationary if it has a Killing field that is everywhere timelike. (M, gab ) is static if it has a Killing field that is everywhere timelike and locally hypersurface orthogonal. (M, gab ) is homogeneous if its Killing fields, at every point of M , span the tangent space. (We shall have another example in section 3.2, where we consider “stationary, axi-symmetric spacetimes”.) The distinction between stationary and static spacetimes should be clear from our discussion in the preceding section. (Recall proposition 2.8.1.) Roughly speaking, in a stationary spacetime there is, at least locally, a “timelike flow” that preserves all spacetime distances. But the flow can exhibit rotation. Think of a whirlpool. It is the latter possibility that is ruled out when one passes to a static spacetime. For example, G¨ odel spacetime, as we shall see, is stationary but not static. Problem 2.9.1. Let κa be a timelike Killing field that is locally hypersurface orthogonal (κ[a ∇b κc] = 0).

Further, let κ be the length of κa . (So κ2 = κn κn .) Show that

κ2 ∇a κb = − κ[a ∇b] κ2 . By way of example, let us find all Killing fields on Minkowski spacetime. This will be easy, as much of the work has already been prepared in sections 1.9 and 2.6. Let κa be a Killing field on Minkowski spacetime (M, gab ). Arguing exactly as in proposition 1.9.9, we can show that, given any point p in M , there is a unique constant, anti-symmetric field Fab on M and a unique constant field k a on M such that κb = χa Fab + kb ,

(2.9.1)

where χa is the position field relative to p. (Recall that Fab = ∇a κb , and kb = κb − χa Fab .) Thus there

is a one-to-one correspondence between Killing fields on Minkowski spacetime and pairs (Fab , kb ) at any 22 They

are “local” because Killing fields need not be complete, and their associated local flow maps need not be defined

globally. (Recall our discussion at the end of section 1.3.)

154


one point, where Fab is an anti-symmetric tensor there and k a is a vector there. It follows that the vector space of Killing fields on Minkowski spacetime has 6 + 4 = 10 dimensions. We can further analyze Fab as in section 2.6. Let ǫabcd be a volume element on M ; let ξ a be a constant, future-directed, unit timelike field on M ; and let E a and B a be defined as in (2.6.9) and (2.6.10): Ea Ba

= F a b ξb, 1 abcd = ǫ ξb Fcd . 2

Then E a and B a are constant fields everywhere orthogonal to ξ a . And it follows from (2.6.13) that we can express κa in the form κb = χa (2 E[a ξb] + ǫabcd ξ c B d ) + kb .

(2.9.2)

This gives us a classification of all Killing fields (relative to an arbitrary choice of “origin” p and constant, unit timelike field ξ a ). Killing fields of the form κb = k b generate (timelike, spacelike, or null) translations. Those of the form κb = χa ǫabcd ξ c B d generate spatial rotations, based at p, with rotational axis B a . Those of the form κb = 2χa E[a ξb] generate boosts, based at p, in the plane determined by ξ a and Ea. Problem 2.9.2. Consider a non-trivial boost Killing field κb = 2χa E[a ξb] on Minkowski spacetime (as determined relative to some point p and some constant unit timelike field ξ a ). “Non-trivial” here means that E a 6= 0. Let η a be a constant field on Minkowski spacetime. Show that £κ η a = 0 iff η a is orthogonal

to both to ξ a and E a . (It follows that the boost isometries generated by κa leave intact all two-dimensional submanifolds orthogonal to ξ a and E a , but “rotate” all two-dimensional submanifolds to which ξ a and E a are tangent.) Problem 2.9.3. This time, consider a non-trivial rotational Killing field κb

=

χa ǫabcd ξ c B d on

Minkowski spacetime (with B a 6= 0). Again, let η a be a constant field on Minkowski spacetime. Show that

£κ η a = 0 iff η a is a linear combination of ξ a and B a . (It follows that the isometries generated by κa

“rotate” all two-dimensional submanifolds orthogonal to ξ a and B a , but leave intact all two-dimensional submanifolds to which ξ a and B a are tangent.) Now we briefly consider two types of conserved quantity. One is an attribute of point particles with positive mass, the other of extended bodies. Let κa be a Killing field in an arbitrary spacetime (M, gab ) (not necessarily Minkowski spacetime), and let γ : I → M be a smooth, future-directed, timelike curve, with unit tangent field ξ a . We take its image to represent the worldline of a point particle with mass

m > 0. Consider the quantity J = (P a κa ), where P a = m ξ a is the four-momentum of the particle. It certainly need not be constant on γ[I]. But it will be if γ is a geodesic. For in that case, ξ n ∇n ξ a = 0

and hence, by (1.9.12),

ξ n ∇n J = m (κa ξ n ∇n ξ a + ξ n ξ a ∇n κa ) = m ξ n ξ a ∇(n κa) = 0.

(2.9.3)


155

Thus, J is constant along the worldlines of free particles of positive mass. We refer to J as the conserved quantity associated with κa . If κa is timelike, we call J the energy of the particle (associated with κa ).23 If it is spacelike, and if its associated flow maps resemble translations,24 we call J the linear momentum of the particle (associated with κa ). Finally, if κa is spacelike, and if its associated flow maps resemble rotations, then we call J the angular momentum of the particle (associated with κa ). It is useful to keep in mind a certain picture that helps one “see” why the angular momentum of free particles (to take that example) is conserved. It involves an analogue of angular momentum in Euclidean plane geometry. Figure 2.9.1 shows a rotational Killing field κa in the Euclidean plane, the image of a geodesic (i.e., a line) L, and the tangent field ξ a to the geodesic. Consider the quantity J = ξ a κa , i.e., the inner product of ξ a with κa , along L. Exactly the same proof as before (of equation (2.9.3)) shows that J is constant along L.25 But here we can better visualize the assertion. Let us temporarily drop indices and write κ · ξ as one would in ordinary Euclidean vector calculus

(rather than ξ a κa ). Let p be the point on L that is closest to the center point where κ vanishes. At

that point, κ is parallel to ξ. As one moves away from p along L, in either direction, the length kκk of κ grows, but the angle ∠(κ, ξ) between the vectors increases as well. It should seem at least plausible

from the picture that the length of the projection of κ onto the line is constant and, hence, that the inner product κ · ξ = cos(∠(κ, ξ)) kκk kξk is constant. That is how to think about the conservation of angular momentum for free particles in relativity theory. It does not matter that in the latter context we are dealing with a Lorentzian metric and allowing for curvature. The claim is still that a certain inner product of vector fields remains constant along a geodesic, and we can still think of that constancy as arising from a compensatory balance of two factors. Let us now turn to the second type of conserved quantity, the one that is an attribute of extended bodies. Let κa be an arbitrary Killing field, and let Tab be the energy-momentum field associated 23 Of

course, one needs to ask what this notion of energy has to do with the one considered in section 2.4. There,

ascriptions of energy to point particles were made relative to individual unit timelike vectors, and the value of the energy at any point was taken to be the inner product of that vector with the particle’s four-momentum vector. We take the present notion of energy to be primary and the earlier one as derived. At least in the context of Minkowski spacetime, one can always extend a unit timelike vector at a point to a constant unit timelike field (which is, of course, a Killing field) and then understand relativization to the vector as relativization to the associated constant field. And perhaps the earlier usage is properly motivated only in spacetimes where individual unit timelike vectors are extendible to constant fields or, at least, to naturally distinguished Killing fields. (Similar remarks apply to components of “linear momentum” in particular directions.) 24 When

one is dealing with Minkowski spacetime, one can assert without ambiguity that a Killing field generates a

“translation”, or a “spatial rotation”, or a “boost”. Things are not always so simple. Still, sometimes a Killing field in a curved spacetime resembles a Killing field on Minkowski spacetime in certain respects, and then the terminology may carry over naturally. For example, in the case of asymptotically flat spacetimes, one can classify Killing fields by their asymptotic behavior. 25 The

mass m played no special role.

156


κa κa

L ξa

p ξa

κa κa

Figure 2.9.1: κa is a rotational Killing field. (It is everywhere orthogonal to a circle radius, and is proportional to it in length.) ξ a is a tangent vector field of constant length on the line L. The inner product between them is constant. (Equivalently, the length of the projection of κa onto the line is constant.)

with some matter field. Assume it satisfies the conservation condition (∇a T ab = 0). Then (T ab κb ) is divergence free: ∇a (T ab κb ) = κb ∇a T ab + T ab ∇a κb = T ab ∇(a κb) = 0.

(2.9.4)

(The second equality follows from the conservation condition and the symmetry of T ab ; the third follows from the fact that κa is a Killing field.) It is natural, then, to apply Stokes’ theorem to the vector field (T ab κb ). Consider a bounded system with aggregate energy-momentum field Tab in an otherwise empty universe. Then there exists a (possibly huge) timelike world tube such that Tab vanishes outside the tube (and vanishes on its boundary). Tab 6= 0 Tab = 0 S2 S1

Figure 2.9.2: The integrated energy (relative to a background timelike Killing field) over the intersection of the world tube with a spacelike hypersurface is independent of the choice of hypersurface. Let S1 and S2 be (non-intersecting) spacelike hypersurfaces that cut the tube as in figure 2.9.2, and let

157


N be the segment of the tube falling between them (with boundaries included). By Stokes’ theorem26 , Z Z (T ab κb ) dSa (T ab κb ) dSa − S1 S2 Z Z ab (T ab κb ) dSa (T κb ) dSa − = S1 ∩ ∂N S2 ∩ ∂N Z Z ab = (T κb ) dSa = ∇a (T ab κb ) dV = 0. Thus, the integral

R

S

∂N

N

ab

(T κb ) dSa is independent of the choice of spacelike hypersurface S intersecting the

world tube, and is, in this sense, a conserved quantity (construed as an attribute of the system confined to the tube). An “early” intersection yields the same value as a “late” one. Again, the character of the background Killing field κa determines our description of the conserved quantity in question. If κa is R timelike, we take S (T ab κb ) dSa to be the aggregate energy of the system (associated with κa ). And so forth.

Let us now continue the discussion that led to (2.9.3) and derive an inequality governing “total integrated acceleration”. Once again, let κa be a Killing field on an arbitrary spacetime (M, gab ), and let γ : I → M be a smooth, future-directed, timelike curve, with unit tangent field ξ a . We take its image to represent the worldline of a point particle with mass m > 0. Again, we consider the quantity J = (P a κa ),

where P a = m ξ a is the four-momentum of the particle. Even without assuming that γ is a geodesic, we have ξ n ∇n J = m (κa ξ n ∇n ξ a + ξ n ξ a ∇n κa ) = m κa ξ n ∇n ξ a .

(2.9.5)

Now let α be the scalar magnitude of the acceleration field, i.e., α2 = −(ξ n ∇n ξ a )(ξ m ∇m ξa ). Then we

have (see problem 2.9.4)

|ξ n ∇n J| ≤ α

p J 2 − m2 (κn κn ).

(2.9.6)

(Of course, if γ is a geodesic, i.e., if α = 0 everywhere, then |ξ n ∇n J| must vanish everywhere as well. So

we recover our earlier result that J is constant in the case of geodesic motion.) If κa is causal (timelike or null) and future-directed everywhere, then J = P a κa > 0, and it follows that |ξ n ∇n J| ≤ α J.

(2.9.7)

So, in this case, the total integrated acceleration of γ — the integral of α with respect to elapsed time — satisfies T A(γ) =

Z

γ

α ds ≥

Z

γ

Z |ξ n ∇n J)| n ds ≥ ξ ∇n (ln J) ds . J γ

(2.9.8)

Thus, if γ passes through points p1 and p2 , the total integrated acceleration between those points is, at least, |(ln J)|p2 − (ln J)|p1 |. (For applications of (2.9.8), see Chakrabarti, Geroch, and Liang [7].) Problem 2.9.4. Derive the inequality (2.9.6). 26 See

Wald [60, Appendix B.2] for a discussion of integration on manifolds and Stokes’ theorem. We did not take the time

to develop these topics in our review of differential geometry because we have so little need of them. This is the only place in this book where reference is made to integration on manifolds (except for the simple case of integration over curves).


2.10

158

The Initial Value Formulation

In this very brief section, we say a few words about the “initial value formulation” of general relativity and make precise the sense in which it is a deterministic theory. (See Hawking and Ellis [30] and Wald [60] for a proper treatment of the subject.) Let S be a smooth, achronal, spacelike hypersurface in our background spacetime (M, gab ). Recall (section 2.5) that D(S), the domain of dependence of S, is the set of all points p in M with this property: given any smooth causal curve without endpoint, if its image passes through p, then it intersects S. Our goal is to explain the sense in which (at least in the empty space case) “what happens on S uniquely determines what happens on D(S)”. Of special interest is the case where S is a Cauchy surface in (M, gab ), i.e., a smooth achronal spacelike hypersurface such that D(S) = M . The first thing we must do is specify what is to count as “initial data” for the metric gab on S. Let ξ a be the (unique) smooth, future-directed, unit timelike field that is everywhere orthogonal to S. (We will refer to it, simply, as the normal field to S.) Our first piece of initial data on S is the induced (negative definite) spatial metric hab = gab − ξa ξb . Our second piece is the extrinsic curvature field πab on S. We 1 can think of the latter as the time derivative of hab in the direction ξ a , at least up to the factor , since 2 2 πab = £ξ hab . (Recall (1.10.17).) Thus our metric initial data on S consists of the pair (hab , πab ), the first and second fundamental forms on S. They correspond, respectively, to position and momentum in the initial value formulation of Newtonian particle mechanics. We know from our discussion in section 1.10 that these fields satisfy a number of constraint equations, including R − (πa a )2 + πab π ab Dc πa c − Da πc c

1 R gab ) ξ a ξ b , 2 = hm a hnp ξ r Rmnpr ,

= −2 (Rab −

where D is the derivative operator induced on S, Rabcd is its associated Riemann curvature field, and R

is the contracted scalar curvature field. (The first equation is just (1.10.21) and we get the second from (1.10.19) by contraction.) Using the symmetries of Rmnpr , we can re-express the right side of the second

equation: hm a hnp ξ r Rmnpr

= =

hm a (g np − ξ n ξ p ) ξ r Rnmrp = hm a ξ r Rmr 1 hm a ξ r (Rmr − R gmr ). 2

And therefore, using Einstein’s equation, we can express our two constraint equations as R − (πa a )2 + πab π ab

= −16 π Tab ξ a ξ b ,

(2.10.1)

Dc πa c − Da πc c

= 8 π Tmr hm a ξ r .

(2.10.2)

159


For simplicity, we shall restrict attention to the empty space case — where Tab vanishes and it is only the evolution of the metric field gab itself that we need to consider. In this special case, of course, the constraint equations assume the form R − (πa a )2 + πab π ab

= 0,

(2.10.3)

Dc πa c − Da πc c

= 0.

(2.10.4)

We started with a spacetime (M, gab ) and moved to an induced initial data set (hab , πab ) on a smooth, achronal, spacelike hypersurface S in M satisfying particular constraint equations. Now we reverse direction. We need a few definitions. Let us say officially that an (empty space) initial data set is a triple ˜ ab , π ˜ ab is a smooth negative(Σ, h ãb ) where Σ is a smooth, connected, three-dimensional manifold, h definite metric on Σ, π ãb is a smooth symmetric field on Σ, and the latter two satisfy the constraint equations (2.10.3) and (2.10.4). ˜ ab , π A Cauchy development of such an initial data set (Σ, h ãb ) is a triple ((M, gab ), S, ϕ) where (i) (M, gab ) is a spacetime that satisfies the field equation Rab = 0, (ii) S is a Cauchy surface in M , (iii) ϕ ˜ ab = φ∗ (hab ) and π is a diffeomorphism of Σ onto S, and (iv) h ãb = φ∗ (πab ), where hab and πab are the first and second fundamental forms induced on S. A Cauchy development ((M, gab ), S, ϕ) of (Σ, ˜hab , π ãb ) is maximal if, in addition, given any other ′ ′ ′ ′ ˜ ãb ), there is an isometry ψ of M ′ into M that respects Cauchy development ((M , g ), S , ϕ ) of (Σ, hab , π ab

′

Σ in the sense that ψ ◦ ϕ = ϕ. Our basic result (due to Choquet-Bruhat and Geroch [8]) is the following. Proposition 2.10.1. Every empty space initial data set has a maximal Cauchy development. It is unique ′ in the following sense. If ((M, gab ), S, ϕ) and ((M ′ , gab ), S ′ , ϕ′ ) are both maximal Cauchy developments ˜ ab , π of (Σ, h ãb ), there is a diffeomorphism ψ : M ′ → M such that ψ ◦ ϕ′ = ϕ and g ′ = ψ ∗ (gab ). ab

Proposition 2.10.1 makes precise the sense in which general relativity is a deterministic theory. But that sense is local in character because it need not be the case in an arbitrary spacetime (M, gab ) that there is any one achronal spacelike hypersurface S such that D(S) = M , i.e., it need not be case that there is a Cauchy surface. (For example, the spacetime that arises by taking the universal covering space of anti-deSitter spacetime admits no Cauchy surface. See Hawking and Ellis [30], section 5.2.)

160


2.11

Friedmann Spacetimes

In this section, we briefly consider the class of Friedmann (or Friedmann-Lemaˆıtre-Robertson-Walker) spacetimes. These are the “standard models” of relativistic cosmology. (For a more complete discussion, see Wald [60] or almost any text in general relativity.) We include this section, even though we are not otherwise undertaking to survey known exact solutions to Einstein’s equation, because we have a particular interest in comparing relativistic cosmology with Newtonian cosmology. We consider the latter in section 4.4. We take a Friedmann spacetime to be one that satisfies a particular symmetry condition — “spatial homogeneity and isotropy” — together with supplemental constraints in the form of energy conditions and/or equations of state. We start with the symmetry condition. Roughly speaking, a spacetime is spatially homogeneous and isotropic if there is a congruence of timelike curves filling the spacetime such that “space”, as determined relative to the congruence, “is the same in all directions”. Here is one way to make the condition precise. (We opt for a local version of the condition. And to avoid certain distracting complications, we cast the definition directly in terms of the existence of isometries, rather than in terms of Killing fields as we did with several symmetry conditions at the beginning of section 2.9.) Let (M, gab ) be a spacetime, and let ξ a be a smooth, future-directed, unit timelike field on M that is twist-free, i.e., ξ[a ∇b ξc] = 0. (So, at least locally, it is possible to foliate M with a one-parameter family

of spacelike hypersurfaces that are orthogonal to ξ a . Recall our discussion in section 2.8. We can think of each of these hypersurfaces as constituting “space” at a given time relative to ξ a .) We say that (M, gab ) is spatially homogeneous and isotropic relative to ξ a if, for all points p in M , and all unit spacelike vectors 1

2

σ a and σ a at p that are orthogonal to ξ a , there is an open set O containing p and an isometry ϕ : O → O 1

2

that keeps p fixed, preserves the field ξ a , and maps σ a to σ a (i.e., such that ϕ(p) = p, ϕ∗ (ξ a ) = ξ a , and 1

2

ϕ∗ (σ a ) = σ a ).27 We further say that (M, gab ) is spatially homogeneous and isotropic if it is so relative to some choice of ξ a . The strength of the condition will become clear as we proceed. We assume in what follows that ξ a is as in the preceding paragraph and (M, gab ) is spatially homogeneous and isotropic relative to ξ a . We first abstract a few general principles. 27 Note,

we require here that ϕ map the field ξ a onto itself everywhere, not just at p. If we required only that it keep

fixed the vector ξ a |p , the condition would not be strong enough for our purposes. For example, Minkowski spacetime would then qualify as spatially homogeneous and isotropic relative to any smooth, future-directed, unit timelike vector field ξ a that is twist-free. It would not have to be the case, as we want it to be, that hypersurfaces orthogonal to ξ a are manifolds of constant curvature. For the corresponding global version of the condition, we would require at the outset that ξ a be (globally) hypersurface 1

2

orthogonal and require that, for all p, σa , and σ a as specified, there is a (global) isometry ϕ : M → M that keeps p fixed, 1

2

preserves the field ξ a , and maps σ a to σ a . We shall later consider what turns on the difference between these two (local vs. global) versions of the spatial homogeneity and isotropy condition.

161


(1) Given any field λa on M , if it is definable in terms of, or otherwise determined by, gab and ξ a , then it must be proportional to ξ a . (So λa = λ ξ a where λ = λn ξ n . And if λa is also orthogonal to ξ a , then λa = 0.) This follows, for if at some point p, λa had a non-zero component orthogonal to ξ a , it would determine a “preferred” orthogonal direction there and violate the isotropy condition. (Here is the argument in more detail. Since that component is determined by gab and ξ a , it must be invariant under all maps that preserve gab and ξ a and that leave p fixed. But, by our assumption of spatial homogeneity and isotropy, the only vector at p, orthogonal to ξ a , that is invariant under all such maps is the zero vector.) It follows from (1), for example, that the acceleration field ξ n ∇n ξ a must vanish, i.e., ξ a must be a geodesic field. (2) Given any scalar field λ on M , if it is definable in terms of, or otherwise determined by, gab and ξ a , then it must be constant on all spacelike hypersurfaces orthogonal to ξ a . (So ∇a λ = (ξ n ∇n λ) ξa .) This is an immediate consequence of (1) as applied to hab ∇b λ, where hab is the spatial projection field (gab − ξa ξb ). So, for example, we have

∇a θ = (ξ n ∇n θ) ξa ,

(2.11.1)

where θ = ∇m ξ m . (Recall section 2.8.) (3) Given any symmetric field λab on M , if it is definable in terms of, or otherwise determined by, gab and ξ a , then it must be of the form λab = α ξa ξb + β hab for some scalar fields α and β. (And if λab is also orthogonal to ξ a , then it must be of the form λab = β hab .) To see this, consider any point p. By (1) as applied to λa b ξ b , there is a number α such that λa b ξ b = α ξ a at p. Now consider the tensor (λab − α ξ a ξ b ) at p. It is symmetric and orthogonal to ξ a in both

indices. So we can express it in the form

1

1 1

2 2

2

3 3

3

λab − α ξ a ξ b = −(σ σ a σ b + σ σ a σ b + σ σ a σ b ), 1

(2.11.2)

3

where the vectors σ a , ..., σ a , together with ξ a , form an orthonormal (eigen)basis for gab at p. But now, 1

2

3

by the isotropy condition, the coefficients σ, σ, σ must be equal. (For all i and j, there is an isometry i

j

that leaves p and (λab − α ξ a ξ b ) fixed but takes σ a to σ a .) If their common value is β, then the right side tensor in 2.11.2 can be expressed as β hab .

It follows from (3) that the shear tensor field σab associated with ξ a must be of the form σab = β hab . But σab is “trace-free”, so 0 = σa a = 3 β. Thus, ξ a has vanishing shear in addition to being geodesic. And we assumed at the outset that it is twist-free. So, by (2.8.13), ∇a ξb =

1 hab θ. 3

(2.11.3)

162


It also follows from (3) that we can construe (M, gab ) as an exact solution to Einstein’s equation for a perfect fluid source with four velocity ξ a . For if Rab = α ξa ξb + β hab , then Rab −

1 Rgab = 8 π(ρ ξa ξb − p hab ), 2

(2.11.4)

(α + β) (α − 3 β) and p = . (This perfect fluid need not satisfy any of the standard energy 16π 16π conditions. We shall soon add one of those conditions as a supplemental constraint, but will work where ρ =

without it for now.) In what follows, we take Tab to be the indicated energy-momentum field, i.e., we take Tab = ρ ξa ξb − p hab . So (after inversion of (2.11.4)), Rab = 8 π(Tab −

1 T gab ) = 4 π(ρ + 3p) ξa ξb − 4 π(ρ − p)hab . 2

(2.11.5)

Next we consider the geometry of spacelike hypersurfaces orthogonal to ξ a . Let S be one such hyper-

surface, and let hab and πab be the first and second fundamental forms induced on S. (Recall section 1.10.) 1 Note that, by (1.10.16) and (2.11.3), the latter assumes a simple form: πab = ha m hb n ∇m ξn = hab θ. 3 Now let Ra bcd be the curvature field associated with the induced derivative operator D. Our goal is to derive an expression for Ra bcd in terms of θ, ρ, and p. We do so by first deriving one for Rbc and then

invoking a general fact about the relation between the two fields that holds in the special case of three dimensional manifolds. It follows from (1.10.20), our expression above for πab , and from (2.11.5) that Rbc

= πn n πbc − πab π ac + hnb hpc Rnp − Rmbcr ξ m ξ r 1 1 2 θ hbc − θ2 hbc − 4 π(ρ − p)hbc − Rmbcr ξ m ξ r . = 3 9

So we need only derive an expression for the fourth term on the right side. (Here and in what follows we shall use the abbreviation θ˙ = ξ n ∇n θ.) Note that by (2.11.3) and (2.11.1), ∇c ∇r ξb

= = =

1 1 ∇c (hrb θ) = [ hrb ξc θ˙ + θ ∇c (grb − ξr ξb ) ] 3 3 1 ˙ [ hrb ξc θ − θ ξr ∇c ξb − θ ξb ∇c ξr ] 3 1 1 1 [ hrb ξc θ˙ − θ2 ξr hcb − θ2 ξb hcr ]. 3 3 3

Hence Rmbcr ξ m = 2 ∇[c ∇r] ξb = and, therefore,

1 2 ξ[c hr]b ( θ˙ + θ2 ) 3 3

1 1 Rmbcr ξ m ξ r = − hbc ( θ˙ + θ2 ). 3 3

Substituting this into our expression for Rbc yields Rbc = K hbc ,

(2.11.6)

163


where K =

1 2 3 (θ

˙ − 4 π (ρ − p). Now we invoke our general fact. In the special case of a three+ θ)

dimensional manifold with metric, we (always) have28

Rabcd = (hbc Rad + had Rbc − hac Rbd − hbd Rac ) +

1 (hac hbd − had hbc ) R. 2

(2.11.7)

So it follows from (2.11.6) (and its contracted form R = 3K) that Rabcd = (

K R − 2K) (hac hbd − had hbc ) = (had hbc − hac hbd ). 2 2

(2.11.8)

Thus, recalling our discussion at the end of section 1.9, we see that (S, hab ) has constant curvature K/2. (We shall soon have a more instructive expression for K.)

Now we turn to considerations of dynamics. We claim that θ˙

=

−4 π(ρ + 3 p) −

ρ˙

=

−(ρ + p) θ.

1 2 θ , 3

(2.11.9) (2.11.10)

(We shall continue to use the dot notation. Here ρ˙ = ξ n ∇n ρ.) We get the first from Raychaudhuri’s

equation (2.8.17), using (2.11.5) and the fact that ξ a is geodesic, irrotational, and shear-free. The second is the continuity condition (2.5.5). Recall that the latter follows from the conservation condition ∇a T ab = 0

as applied to our energy-momentum field T ab = ρ ξa ξb − p hab . (And the conservation condition itself is

a consequence of Einstein’s equation.)

It is convenient and customary to introduce a new field a that we can think of as a “scaling factor”. We want it to be constant on spacelike hypersurfaces orthogonal to ξ a , i.e., hmn ∇m a = 0. So we need

only specify its growth along any one integral curve of ξ a . We define it, up to a multiplicative constant, by the condition

a˙ 1 θ = . 3 a

(2.11.11)

(Certainly this equation has solutions. Indeed, if the curve is parametrized by a time function t where Z t θ ξa = ∇a t, then all functions of the form a(t) = ef (t) , with f (t) = dt, qualify.) The condition t0 3 inherits a natural interpretation from the one we have given for θ. It concerns the rate of volume increase for a fluid with four-velocity ξ a . We saw in section 2.8 that if an (“infinitesimal”) blob of the fluid has volume V , then V˙ = V θ. (Recall (2.8.15).) If we think of the blob as a cube whose edges have length a, then V = a3 and we are led immediately to (2.11.11). It is in this sense that a is a scaling factor. If we now express our equations for θ˙ and ρ˙ above in terms of a, we have 3

28 We

a ¨ a

= −4 π(ρ + 3 p),

ρ˙

= −3

a˙ (ρ + p) a

(2.11.12) (2.11.13)

shall later prove a close analogue of this result (proposition 4.1.4) in connection with our discussion of classical

spacetimes. It should be clear how to adapt the proof to the present context. (We present the argument there rather than here because of added complications that arise when one is dealing with classical spacetimes.)

164


where, of course, a ¨ = ξ n ∇n (ξ m ∇m a). These two jointly imply (by integration) that there is a number k

such that

a˙ 8π k ( )2 − ρ = − 2. a 3 a

(2.11.14)

(This is “Friedmann’s equation”.) Since a was only determined initially up to a multiplicative constant, we can now normalize it so that k = −1 or k = 0 or k = 1. We can use the listed equations to express several fields of interest directly in terms of the scaling factor a and k: 8πρ

=

a˙ k 3 ( )2 + 3 2 , a a

8πp

=

−2

Rabcd

=

−

a ¨ a˙ k − ( )2 − 2 , a a a

k (had hbc − hac hbd ). a2

(2.11.15) (2.11.16) (2.11.17)

Here (2.11.15) is just a reformulation of (2.11.14). (2.11.16) follows from (2.11.12) and (2.11.15). For K 1 ˙ − 4 π (ρ − p). (2.11.17), recall that, by (2.11.8), Rabcd = (had hbc − hac hbd ), where K = (θ2 + θ) 2 3 But ¨ a˙ 1 2 ˙ = a (θ + θ) + 2 ( )2 3 a a by (2.11.9), (2.11.11), and (2.11.12). And it follows from (2.11.15) and (2.11.16) that 4 π (ρ − p) =

a˙ k a ¨ + 2 ( )2 + 2 2 a a a

k , as claimed. a2 (2.11.17) tells us that (S, hab ) has constant curvature −k/a2 . Remember, though, that hab is negative

So K = −2

definite, and curvature is usually reported in terms of the positive definite metric −hab . This introduces

a sign change. (The switch from hab to −hab leaves D, Ra bcd , and (had hbc − hac hbd ) intact, but reverses

the sign of Rabcd = han Rn bcd .) So we shall record our conclusion this way:

(S, −hab ) is a manifold of constant curvature, and the magnitude of its curvature is (−1/a2 ),

0, or (1/a2 ) depending on whether k is −1, 0, or 1.

We have reached this point assuming only a local version of the spatial isotropy condition. But now suppose for a moment that the global version holds as well, and let S be any maximally extended spacelike

hypersurface that is everywhere orthogonal to ξ a . Then we can say more about the global structure of (S, −hab ). In this case, it follows from the way the global condition is formulated that (S, −hab ) is, itself, 1

2

a homogeneous, isotropic three-manifold in this sense: for all points p in S, and all unit vectors σ a and σ a 1

2

in the tangent space to S at p, there is an isometry ψ : S → S that keeps p fixed and that maps σ a to σ a .

This is a very strong constraint and rules out all but a small number of possibilties (Wolf [64]). If k = 0, (S, −hab ) cannot be just any flat three-manifold. It must be isometric to three-dimensional Euclidean

165


space, i.e., it must also be diffeomorphic to R3 and geodesically complete.29 If k = −1, (S, −hab ) must

be isometric to three-dimensional hyperbolic space H 3 . (We shall return to consider one realization of three-dimensional hyperbolic space at the end of the section.) Finally, if k = 1, (S, −hab ) must be

isometric either to three-dimensional spherical space S 3 or to three-dimensional elliptic three-space P 3 . The latter arises if one identifies “antipodal points” in the former. Let us now revert to the local version of the spatial homogeneity and isotropy condition — leaving open the global structure of maximally extended spacelike hypersurface orthogonal to ξ a — and continue with our consideration of dynamics. The difference in strength between the two versions of the condition plays no role here. So far, assuming only the spatial homogeneity and isotropy condition, we have established that the scaling function a must satisfy (2.11.15) and (2.11.16). Now for the first time, just so as to have one example, we assume that our perfect fluid satisfies a particular equation of state, namely p = 0, and consider how the latter constrains the growth of the scaling function. (We are certainly not claiming that this assumption is realistic, i.e., holds (approximately) in our universe.) If we insert this value for p in (2.11.16) and multiply by a2 a, ˙ we arrive at 2 a ¨ a˙ a + a˙ 3 + k a˙ = 0. It 8π follows that there is a number C such that a˙ 2 a + k a = ρ a3 = C. (The first equality follows from 3 (2.11.15).) So our task is now reduced to solving the differential equation a˙ 2 −

C + k = 0. a

(2.11.18)

The solutions are the following. (It is convenient to express two of them in parametric form.)

k

k

k

=

=

=

−1

0

+1

    a(x)    t(x) a(t) =     a(x)    t(x)

C (cosh x − 1) 2 C (sinh x − x) 2

x ∈ (0, ∞)

9C 4

13

t ∈ (0, ∞)

=

C (1 − cos x) 2

=

C (x − sin x) 2

= =

2

t3

x ∈ (0, 2 π)

These are maximally extended solutions for the case where θ is positive at at least one point. We get additional (time-reversed) solutions if we assume that θ is negative at at least one point. 29 This

should seem, at least, intuitively plausible. Consider a lower dimensional case. The Euclidean plane is not the

only two-dimensional Riemannian manifold of constant 0 curvature. The cylinder and the torus also qualify. But neither of them is isotropic in the relevant sense. For given a point in either, the only global isometry of the manifold that keeps the point fixed is the identity map.

166


Rough (qualitative) graphs of these solutions are given in figure 2.11.1. If k = −1 or k = 0, expansion da starts at the big bang and continues forever. In both cases, the rate of expansion decreases monotondt ically. But there is this difference: the rate of expansion shrinks to 0 asymptotically when k = 0, but has a limit value that is strictly positive when k = −1. (One curve is asymptotically flat; the other is not.) Cπ ) and In contrast, if k = 1, expansion continues until a maximum value is reached for a (at time t = 2 then a period of accelerating contraction begins that leads to a big crunch. a

k = -1

k=0

k=1

t

Figure 2.11.1: Rough graphs of the scaling factor a in the three cases.

Problem 2.11.1. Confirm that the three stated solutions do, in fact, satisfy (2.11.18). Problem 2.11.2. Consider a second equation of state, namely that in which ρ = 3 p. (For Tab = ρ ξa ξb − p hab , this is equivalent to T = 0.) Show that in this case there is a number C ′ such that a˙ 2 a2 + k a2 =

8π 4 ρ a = C′. 3

(So in this case, the equation to solve is not (2.11.18), but rather a˙ 2 −

C′ + k = 0.) a2

It will be instructive to consider an ultra-simple, degenerate Friedmann spacetime and see how some of our claims turn out in this special case. Let (M, gab ) be Minkowski spacetime. Let o be any point in M , and let O be the (open) set of all points p in M such that o ≪ p, i.e., such that there is a smooth

future-directed timelike curve that runs from o to p. (See figure 2.11.2.) Further, let χa be the position field based at o — so χa vanishes at o and ∇a χb = 0 — and let ξ a be the field 1

ξ a = (χb χb )− 2 χa as restricted to O. The latter is, clearly, a smooth, future-directed, unit timelike field on O. Moreover, it is (globally) hypersurface orthgonal, i.e., there exist smooth scalar fields f and g on O such that

167


1

ξa = f ∇a g. Indeed, if χ = (χa χa ) 2 , then χa = χ ξ a , and ∇n χ =

1 1 1 (χa χa )− 2 ∇n (χb χb ) = (χa χa )− 2 χb δn b = χ−1 χn = ξn . 2

(2.11.19)

We claim that the restricted spacetime (O, gab|O ) is spatially homogeneous and isotropic with respect to ξ a and so qualifies as a Friedmann spacetime (with ρ = p = 0). Indeed, this reduces to a standard claim about the symmetries of Minkowski spacetime. Given any point p in O, and any two (distinct) 1

2

unit spacelike vectors σ a and σ a at p that are orthogonal to ξ a , there is a spatial rotation that keeps p 1

2

fixed, preserves the field ξ a , and takes σ a to σ a .30

o Figure 2.11.2: Minkowski spacetime (in profile) as restricted to the set of all points to the timelike future of a point o. It qualifies as a (degenerate) Friedmann spacetime with ρ = p = 0. A χ = constant hyperboloid is indicated. It (together with the metric induced on it) is a realization of three-dimensional hyperbolic space.

We know from our earlier discussion that (2.11.3) must hold. In this special case, it is easy to check the result with a direct computation. By (2.11.19), we have θ = ∇a ξ a = ∇a (χ−1 χa ) = χ−1 (∇a χa ) − χ−2 χa ∇a χ = 4 χ−1 − χ−1 = 3 χ−1 ,

(2.11.20)

and, hence, ∇a ξb

∇a (χ−1 χb ) = χ−1 (∇a χb ) − χ−2 χb ∇a χ = χ−1 gab − χ−2 χb ξa 1 = χ−1 gab − χ−1 ξb ξa = χ−1 hab = θ hab , 3 =

as expected. Notice also, that if we take a = χ, then a˙ = ξ a ∇a χ = 1 by (2.11.19) and a˙ 1 θ = . 3 a 30 Let

o

ξa be a constant unit timelike field on O that agrees with ξ a at p, and let σa be a constant unit spacelike field 1

2

that is orthogonal to all three vectors ξ a , σa , and σa at p. Then the rotation in question is generated by the Killing field o

κb = ǫabcd χa ξ c σd (for either choice of volume element ǫabcd ). Recall (2.9.2).

168


This choice of a satisfies Friedmann’s equation (2.11.14) with ρ = 0 and k = −1. Now consider the hyperboloids in O defined by the condition χ = constant. (See figure 2.11.2 again.) Each is a spacelike hypersurface that is everywhere orthogonal to ξ a . (For if σ a is a vector at a point of one such hypersurface S that is tangent to S, then σ n ∇n χ = 0 and, therefore, by (2.11.19), σ n ξn = σ n ∇n χ = 0.)

Let S be one such hyperboloid. Let D be the induced derivative operator on S, and let Rabcd be its

associated curvature field. We know from (2.11.17) that Rabcd =

1 (had hbc − hac hbd ), χ2

since here a = χ and k = −1. Again, we can check this directly. To do so, we first compute the second

fundamental form πab on S. (Recall (1.10.16).) Since hb n χn = hb n (χ ξn ) = 0, we have

πab = ha m hb n ∇m ξn = ha m hb n ∇m (χ−1 χn ) = χ−1 ha m hb n ∇m χn = χ−1 ha m hb n gmn = χ−1 hab . It follows, by (1.10.22), that Rabcd = πad πbc − πac πbd =

1 (had hbc − hac hbd ), χ2

as expected. Thus, if S is characterized by the value χ, then (S, −hab ) is a three-dimensional manifold with constant

curvature −1/χ2 . Moreover, as we know from our discussion above, it cannot be just any such manifold,

but must be, in fact, isometric to three-dimensional hyperbolic space H 3 . If we had started with a three-

dimensional version of Minkowski spacetime, our hyperboloid (with induced metric) would be isometric to two-dimensional hyperbolic space, otherwise known as the Lobatchevskian plane. (For more about this “hyperboloid model” for Lobatchevskian plane geometry see, e.g., Reynolds [52].) Finally, recall the remarks we made in section 2.7 about the cosmological constant Λ. If we include the constant in Einstein’s equation, i.e., if we take the latter to be (2.7.4), then Raychaudhuri’s equation (2.8.17) yields

1 θ˙ = −4 π(ρ + 3 p) − θ2 + Λ 3

(2.11.21)

rather than (2.11.9). This, in turn, leads to Friedmann’s equation in the form Λ 8π k a˙ ρ = − 2 + ( )2 − a 3 a 3

(2.11.22)

rather than (2.11.14). Equation (2.11.21) serves to explain Einstein’s introduction of the cosmological constant. He thought he needed to find a non-expanding model (θ = 0) to represent the universe properly. And, in our terms, he was considering only Friedmann spacetimes and only perfect fluid sources that are pressureless (p = 0) and non-trivial (ρ > 0). It is an immediate consequence of (2.11.21) that these conditions can be satisfied


169

if, but only if, Λ = 4πρ > 0. And in this case, it follows from (2.11.22) and (2.11.11) that k/a2 = 4πρ. So (since k is normalized to be 1, 0, or −1), we see that the stated conditions can be satisfied iff Λ 1 a2 k

=

4πρ > 0

=

4πρ

=

1.

(These conditions characterize Einstein static spacetime.) It is also an immediate consequence of (2.11.21) — at least if our universe can be represented as a Friedmann spacetime — that evidence for an accelerating rate of cosmic expansion (θ˙ > 0) counts as evidence either for a positive value for Λ or for a violation of the strong energy condition. (Recall from problem 2.7.3 that a perfect fluid satisfies the strong energy condition iff ρ + p ≥ 0 and ρ + 3 p ≥ 0.)

Chapter 3

Special Topics 3.1

G¨ odel Spacetime

Kurt G¨ odel is, of course, best known for his work in mathematical logic and the foundations of mathematics. But in the late 1940s he made an important contribution to relativity theory by finding a new solution to Einstein’s equation (G¨ odel [25]). It represents a possible universe with remarkable properties. For one thing, the entire material content of the Gödel universe (on a cosmological scale) is in a state of uniform, rigid rotation. For another, light rays and free test particles in it exhibit a kind of boomerang effect. Most striking of all, the Gödel universe allows for the possibility of “time travel” in a certain interesting sense.1 Though not a live candidate for describing our universe (the real one), Gödel’s solution is of interest because of what it tells us about the possibilities allowed by relativity theory. In this section, we present the solution and establish several of its basic properties in a running list. We shall later use it as an example when we consider orbital rotation in section 3.2. It will be helpful to keep in mind two different coordinate expressions for the Gödel metric and also a coordinate-free characterization. We start with the former. Let us officially take Gödel spacetime to be the pair (M, gab ), where M is the manifold R4 and where e2x x 2 (da y)(db y) − (da z)(db z) + 2 e (d(a t)(db) y) . gab = µ (da t)(db t) − (da x)(db x) + 2 1 In

(3.1.1)

addition to finding this one new exact solution to Einstein’s equation, G¨ odel [26] also established the existence of

solutions representing universes that are rotating and expanding, though he did not exhibit any of the latter explicitly. For a review of G¨ odel’s contributions to relativity theory and cosmology (and subsequent work on rotating solutions), see Ellis [18].

170

171

CHAPTER 3. SPECIAL TOPICS

Here µ is an arbitrary positive number (a scale factor), and t, x, y, z are global coordinates on M .2 In what follows, we use the abbreviations a a ∂ ∂ ta = xa = ∂t ∂x

ya =

∂ ∂y

a

za =

∂ ∂z

To confirm that gab is a metric of signature (1, 3), it suffices to check that the fields √ ta xa za 2 a , , (t − e−x y a ), µ µ µ µ

a

.

(3.1.2)

(3.1.3)

form an orthonormal basis (of the appropriate type) at each point. The first, in particular, is a smooth, unit timelike vector field on M . That there exists such a field shows us that Gödel spacetime is temporally orientable. It is also orientable since the anti-symmetrized product of the four fields in (3.1.3) qualifies as a volume element. (1) G¨ odel spacetime is temporally orientable and orientable. We shall work with the temporal orientation determined by ta in what follows. We note for future reference that the inverse field of gab is g bc =

i 1 h b c b c −2x b c b c −x (b c) , −t t − x x − 2 e y y − z z + 4 e t y µ2

(3.1.4)

and that lowering indices in (3.1.2) with gab yields: ta

= µ2 (∇a t + ex ∇a y),

(3.1.5)

xa

= −µ2 ∇a x, 2x e = µ2 ∇a y + ex ∇a t , 2

(3.1.6)

ya za

= −µ2 ∇a z.

(3.1.7) (3.1.8)

(Here ∇ is the derivative operator on M compatible with gab , and we have switched from writing, for example., “da t” to “∇a t”.)

We claim, first, that the four fields ta , 2 More

ζ a = (xa − y y a ),

ya,

za

(3.1.9)

precisely, t, x, y, z are real-valued functions on M , and the composite map Φ : p 7→ (t(p), x(p), y(p), z(p)) is a

bijection between M and R4 that belongs to the collection C of 4-charts that defines the manifold R4 . The coordinates ∂ t, x, y, z correspond to u1 , u2 , u3 , u4 in the notation of section 1.2. So, for example, we understand the vector ( )a at any ∂t point p to be the tangent there to the curve r 7→ Φ−1 (t(p) + r, x(p), y(p), z(p)).

172


are all Killing fields. They are, in fact, the generators, respectively, of one-parameter (global) isometry ζ

t

y

z

groups {Γr }r∈R , {Γr }r∈R , {Γr }r∈R , {Γr }r∈R on M defined by: t

Γr (p) ζ

Γr (p) y

Γr (p) z

Γr (p)

= Φ−1 (t(p) + r, x(p), y(p), z(p)) , = Φ−1 (t(p), x(p) + r, e−r y(p), z(p)), = Φ−1 (t(p), x(p), y(p) + r, z(p)), = Φ−1 (t(p), x(p), y(p), z(p) + r),

where Φ : M → R4 is the chart defined by Φ(p) = (t(p), x(p), y(p), z(p)). Here, of course, the group

operation is composition.3 An equivalent formulation may be more transparent. For example, we can ζ

understand Γr to be defined by the requirement that, for all numbers t0 , x0 , y0 , z0 , ζ

(Φ ◦ Γr ◦ Φ−1 ) (t0 , x0 , y0 , z0 ) = (t0 , x0 + r, e−r y0 , z0 ). The field xa is not a Killing field, but it is the generator of the one-parameter group of diffeomorphisms x

{Γr }r∈R on M given by x

= Φ−1 (t(p), x(p) + r, y(p), z(p)).

Γr (p)

The five fields under consideration satisfy the following Lie bracket relations:

3 There

[ta , ζ a ] = [ta , y a ] = [ta , z a ] = [ζ a , z a ] = [y a , z a ] = 0,

(3.1.10)

[xa , ta ] = [xa , ζ a ] = [xa , y a ] = [xa , z a ] = 0,

(3.1.11)

are a few things that have to be checked. First, each of these maps (for any choice of r) is, in fact, an isometry. ζ

This follows from basic facts we have recorded in section 1.5. Consider Γr , for example. By (1.5.6) and (1.5.7), we have ζ

ζ

ζ

(Γr )∗ (e2x ) = e2(x+r) and (Γr )∗ (da y) = da ((Γr )∗ (y)) = da (e−r y) = e−r (da y). Hence, ζ

ζ

ζ

ζ

(Γr )∗ (e2x (da y)(db y)) = ((Γr )∗ (e2x )) ((Γr )∗ (da y)) ((Γr )∗ (db y)) = e2x (da y)(db y). ζ

ζ

Arguing in this way, we can show that all the terms in gab are preserved by (Γr )∗ and, so, (Γr )∗ (gab ) = gab . Second, each of the groups does, in fact, have the indicated vector field as its generator. This follows from our discussion ζ

in sections 1.2 and 1.3. Consider {Γr }r∈R , for example. Let p be a point with coordinates Φ(p) = (t0 , x0 , y0 , z0 ), and let

γ : R → M be the curve through p defined by

ζ

γ(r) = Γr (p) = Φ−1 (t0 , x0 + r, e−r y0 , z0 ). →

We need to show that γ

a

= ζ a at all points on the image of γ. Let f be any smooth field on some open set containing p.

Then, by the chain rule, at all points γ(r), →a

γ (f )

= =

d d ∂ (f ◦ Φ−1 ) ∂ (f ◦ Φ−1 ) (f ◦ γ) = (f ◦ Φ−1 )(t0 , x0 + r, e−r y0 , z0 ) = ·1 + · (−e−r y0 ) 2 dr dr ∂x ∂ x3 „ « ∂f ∂f ∂f ∂f ∂ a ∂ a ·1 + · (−e−r y0 ) = − y = ( ) − y( ) (f ) = ζ a (f ). ∂x ∂y ∂x ∂y ∂x ∂y

So we are done. Here x1 , x2 , x3 , x4 are the coordinate projection functions on R4 that we considered in section 1.2. So, for ∂ (f ◦ Φ−1 ) ∂f = is an instance of (1.2.7). (As mentioned in the preceding example, (x3 ◦ Φ)(p) = y(p). And the equality ∂y ∂ x3 note, the coordinates t, x, y, z correspond to u1 , u2 , u3 , u4 in the notation of section 1.2.)

173


[ζ a , y a ] = y a .

(3.1.12)

There are various ways to see why these hold. For those in the first two rows, it is easiest to invoke a basic result (that we did not formulate in chapter 1). Proposition 3.1.1. Let αa and β a be smooth fields on a manifold that generate one-parameter groups of β

α

α

β

diffeomorphisms {Γr }r∈R and {Γr }r∈R on that manifold. Then [αa , β a ] = 0 iff Γr and Γr′ commute

for all r and r′ .

(See, for example, Spivak [57, volume 1, p. 217].) It is clear in each case that the relevant commutation ζ

t

relations obtain, e.g., Γr and Γs commute for all r and s.4 For (3.1.12), note that [ζ a , y a ] = −[y a , ζ a ] = −£ya (xa − y y a ) = [xa , y a ] + (£ya y) y a + y [y a , y a ] = y a , since £ya y = y n ∇n y = 1, and [xa , y a ] = [y a , y a ] = 0. t

y

ζ

z

By composing the isometries Γr , Γr , Γr , Γr (with appropriate choices for r in each case), we can go from any one point in M to any other. Moreover, each of the individual isometries, and so any composition of them, preserves the fields ta and z a . (This follows from propositions 1.6.6 and 1.6.4, and the fact that each of the generators ta , ζ a , y a , z a has a vanishing Lie bracket with ta and z a .) So we have the following homogeneity claim. (2) G¨ odel spacetime is (globally) homogeneous in this strong sense: given any two points p and q in M , there is an isometry ψ : M → M such that ψ(p) = q, ψ∗ (ta ) = ta , and ψ∗ (z a ) = z a . (The maps referred to here preserve temporal orientation automatically because they preserve ta , and we are using that field to define temporal orientation.) We shall repeatedly invoke this strong form of homogeneity in what follows. For example, we shall prove an assertion about a particular integral curve of ta (that makes reference only to gab , ta , and za ), and then claim that it necessarily holds for all integral curves of that field. The four Killing fields ta , ζ a , y a , z a are clearly independent of each other. In fact, one can find a fifth that is independent of these four, e.g., κ

a

= =

4 For

a

all p, r, and s, we have t

ζ

Γr (Γs (p))

= =

And a similar computation shows that ζ

t

Γs (Γr (p)) t

1 2 a −2x −2 e t + y x + e − y y 2 1 2 a −x a a −2x −2 e t + y ζ + e + y y . 2 −x a

ζ

ζ

t

So Γr (Γs (p)) = Γs (Γr (p)).

„

« ζ ζ ζ ζ t(Γs (p)) + r, x(Γs (p)), y(Γs (p)), z(Γs (p)) ` ´ Φ−1 t(p) + r, x(p) + s, e−s y(p), z(p) .

Φ−1

=

` ´ Φ−1 t(p) + r, x(p) + s, e−s y(p), z(p) .

174


(To confirm that it is a Killing field, it suffices to expand ∇a κb and use our expressions above for ta , xa ,

and yb to show that its symmetric part vanishes.5 )

Now we do a bit of calculation and derive an expression for the Ricci tensor field Rab . Note first that ∇a tb ∇a xb ∇a yb ∇a zb

= µ2 ex (∇[a x)(∇b] y), 2x 2 e x = µ (∇a y)(∇b y) + e (∇(a y)(∇b) t) , 2 = µ2 e2x (∇[a x)(∇b] y) + ex (∇[a x)(∇b] t) , = 0.

(3.1.13) (3.1.14) (3.1.15) (3.1.16)

These can be checked easily by using (3.1.5) – (3.1.8) and the fact that ta , ζ a , y a , z a are Killing fields. Since ta is a Killing field, for example, we have ∇(a tb) = 0 and, therefore, ∇a tb = ∇[a tb] = µ2 ∇[a ∇b] t + ex ∇[a ∇b] y + ex (∇[a x)(∇b] y) = µ2 ex (∇[a x)(∇b] y).

This gives us (3.1.13). The other cases are handled similarly.6 It follows immediately that ta , xa and z a are all geodesic fields: ta ∇a tb = 0

xa ∇a xb = 0

z a ∇a z b = 0.

(3.1.17)

We shall be particularly interested in the (maximally extended) integral curves of ta . Their images are sets of the form {Φ−1 (t, x0 , y0 , z0 ) : t ∈ R}, for particular choices of x0 , y0 , z0 . We shall call these curves

(or their images) t-lines.

Now we turn to Rab . We claim, first, that symmetry considerations alone establish that it must have the form Rab = α tâ tˆb + β (gab − tâ tˆb − zâ zˆb )

(3.1.18)

where α and β are particular numbers (to be determined), and tâ and zâ are normalized versions of ta and z a . (So ta = µ tâ and z a = µ zâ .) The argument we use to establish this is much like that used in section 2.11 when we considered the Ricci tensor field in Friedmann spacetimes. In both cases, it turns on an isotropy condition. Shortly, when we switch to an alternate coordinate representation of the Gödel metric, it will be clear that given any t-line (through any point), there is a global isometry (a rotation) 5 We

∇a κ b

have =

−2 e−x ∇a tb + 2 e−x (∇a x) tb + y ∇a ζb + (∇a y) ζb + (e−2x +

=

−2 e−x ∇a tb + y ∇a ζb + (e−2x +

1 2 y )∇a yb − 2 e−2x (∇a x) yb + y (∇a y) yb 2

1 2 y )∇a yb + (∇a x)(2 e−x tb − 2 e−2x yb ) + (∇a y)(ζb + y yb ). 2

But (2 e−x tb − 2 e−2x yb ) = µ2 ∇b y and (ζb + y yb ) = −µ2 ∇b x. And the first three terms have vanishing symmetric

part since ta , ζ a , and y a are Killing fields. So ∇(a κb) = 0. 6 For

(3.1.14), note that since ∇[a xb] = −µ2 ∇[a ∇b] x = 0, and since (xa − y y a ) and y a are Killing fields, – » 2x e ∇a xb = ∇(a xb) = y ∇(a yb) + (∇(a y) yb) = (∇(a y) yb) = µ2 (∇a y)(∇b y) + ex (∇(a y)(∇b) t) . 2

175


that leaves fixed every point on the line and also preserves the field z a . In effect, we now make use of that rotational symmetry, but cast the argument in terms of Killing fields rather than of the rotations themselves. Since we can find an isometry that maps any one point in M to any other and preserves both ta and z a , it will suffice to show that (3.1.18) holds at one point, say p = Φ−1 (0, 0, 0, 0). To do so, it will suffice to show, in turn, that the two sides of (3.1.18) yield the same result when contracted with each of the vectors ta , xa , (ta − y a ), z a . (It is convenient to work with this basis because the vectors are mutually

orthogonal at p. It does not matter that they are not normalized.) So our task reduces to showing that the following all hold at p (for some values of α and β). (i) Rab ta = α tb

(ii) Rab xa = β xb

(iii) Rab (ta − y a ) = β (tb − yb )

(iv) Rab z a = 0

Given any Killing field λa in any spacetime, we have Rab λa = Rn abn λa = −Ra nb n λa = ∇n ∇b λn .

(3.1.19)

(The second equality follows from the symmetries of the Riemann curvature tensor field, and the third follows from proposition 1.9.8.) So, in particular, applying this result to the Killing field z a in Gödel spacetime, and recalling (3.1.16), we have Rab z a = ∇n ∇b z n = 0. This gives us (iv). Next, consider the field

1 κ′a = −2 (e−x − 1) ta + y xa + e−2x − y 2 − 1 y a . 2

(3.1.20)

It is a linear combination of Killing fields (κ′a = κa + 2 ta − y a ) and so is, itself, a Killing field. What

is important about it is that it vanishes at p.7 Notice that we have [ta , κ′a ] = [z a , κ′a ] = 0,

(3.1.21)

[xa , κ′a ] = 2 e−x ta − 2 e−2x y a ,

(3.1.22)

[y a , κ′a ] = xa − y y a

(3.1.23)

everywhere8, and so £κ′ xa

= [κ′a , xa ] = −2(ta − y a ),

(3.1.24)

£ κ′ y a

= [κ′a , y a ] = −xa

(3.1.25)

at p. Since κ′a vanishes at p, we have £κ′ f = κ′a ∇a f = 0 at p for all smooth scalar fields f . So, in particular, since κ′ Lie derives Rab (as all Killing fields do) and Lie derives ta (by (3.1.21)), we have

7 It

0

= £κ′ (Rab ta y b ) = Rab ta (£κ′ y b ) = −Rab ta xb ,

(3.1.26)

0

= £κ′ (Rab ta xb ) = Rab ta (£κ′ xb ) = −2 Rab ta (tb − y b )

(3.1.27)

is, in fact, up to a constant, just the rotational Killing field (∂/∂φ)a that we shall consider below. The latter, as we

shall see, generates a one-parameter group of rotations that keep fixed all points on the t-line through p (and preserve z a ). 8 These

all follow easily from the Lie bracket relations that we have already established.

176


at p. These two, together with (iv), show that Rab ta must be proportional to tb at p, which is what we need for (i). Similarly, we have 0 = £κ′ (Rab y a y b ) = Rab £κ′ (y a y b ) = 2 Rab y a (£κ′ y b ) = −2 Rab y a xb at p. This, together with (3.1.26) and (iv), shows that (ii) must hold for some β. Finally, (iii) follows from (ii). For if Rab xa = β xb , then −2 Rab (ta − y a ) = Rab £κ′ xa = £κ′ (Rab xa ) = £κ′ (β xb ) = β £κ′ xb = −2 β (tb − yb ). (For the final equality, we use the fact that £κ′ gab = 0 and, so, £κ′ xb = £κ′ (gab xa ) = gab £κ′ xa = −2 gab (ta − y a ) = −2 (tb − yb ).) Now it only remains to compute α and β in (3.1.18). It follows from (3.1.17) and from (3.1.19) — as applied to the Killing fields ta and xa — that α = =

Rab tâ tˆb = µ−2 Rab ta tb = µ−2 tb ∇n ∇b tn = µ−2 [∇n (tb ∇b tn ) − (∇n tb )(∇b tn )] −µ−2 (∇n tb )(∇b tn )

and (by the same argument) β = −µ−2 (∇n xb )(∇b xn ). Now, raising indices in (3.1.13) and (3.1.14), using (3.1.4), yields ∇n tb

=

∇n xb

=

ex b x (∇n y) + (−e−x y b + tb )(∇n x), 2 ex b t (∇n y) + (−e−x y b + tb )(∇n t) 2

It follows that (∇n tb )(∇b tn ) = −1

(∇n xb )(∇b xn ) = 0

(3.1.28)

and, therefore, α = µ−2 and β = 0. Thus we have Rab = µ−2 tâ tˆb .

(3.1.29)

So R = µ−2 and Rab − Therefore,

1 µ−2 µ−2 ˆ ˆ ta tb − (gab − tâ tˆb ) . R gab = µ−2 tâ tˆb − gab = 2 2 2

(3) G¨ odel spacetime is a solution to Einstein’s equation (without cosmological constant) 1 Rab − R gab = 8 π ρ tâ tˆb − p (gab − tâ tˆb ) 2 for a perfect fluid with four-velocity tâ , mass-density ρ = 1/(16 π µ2 ), and pressure p = 1/(16 π µ2 ). (Equivalently, it is a solution to Einstein’s equation with cosmological constant λ = −1/(2 µ2 ) Rab −

1 R gab − λ gab = 8 πρ′ tâ tˆb 2

for a dust field with mass-density ρ′ = 1/(8 π µ2 ).)

177


Recall that a perfect fluid satisfies the dominant energy condition iff |p| ≤ ρ. So if we construe Gödel

spacetime as a perfect fluid solution to Einstein’s equation without cosmological constant, the perfect fluid in question is only “borderline” for satisfying the condition. Let us further consider the normalized field tâ = ta /µ, which we now understand to represent the

four-velocity of the background source fluid. We know that its associated expansion field θab vanishes (because it is a Killing field), as does its acceleration (by (3.1.17)). Let us now compute its associated rotation field ω a . Let ǫabcd be a volume element on M . (We know that volume elements exist since, for example, t[a xb y c z d] is an anti-symmetric field on M that is everywhere non-vanishing. We need only normalize it to obtain a volume element.) The field ∇a tb is anti-symmetric and it is orthogonal to both ta and z a (by (3.1.13)). So we can express it in the form

∇a tb = f ǫabcd tc z d for some field f . To determine f , we need only contract each side with itself and make use of (3.1.28): 1 = =

(∇a tb ) (∇a tb ) = f 2 ǫabcd tc z d ǫabmn tm zn = −4 f 2 δ [m c δ n] d tc z d tm zn −2 f 2 (tm z n − tn z m ) tm zn = 2 µ4 f 2 .

Taking f to be positive — we can always switch from the volume element ǫabcd to −ǫabcd if necessary —

we have

1 ∇a tb = √ ǫabcd tc z d . 2 2µ

Hence, using this volume element to compute the rotation vector field, ωa

= =

1 abcd ˆ 1 1 abcd ǫabcd tb ǫcdmn tm z n ǫ tb ∇c td = √ ǫ tb ∇c tˆd = 2 2 2µ 2 2 µ4 −4 1 1 √ δ [a m δ b] n tb tm z n = √ (tb tb ) z a = √ z a. 4 4 2 2µ 2µ 2 µ2

(3.1.30)

Let us record this result too. (4) The four-velocity tâ in G¨ odel spacetime is expansion free (θ = 0), shear free (σab = 0), and geodesic n a (tˆ ∇n tˆ = 0), but its rotation field ω a is non-vanishing and constant (∇a ω b = 0). Indeed, ω a is √ just (1/ 2 µ2 ) z a . The G¨ odel universe is thus in a state of uniform, rigid rotation. It turns out that there are only two homogeneous perfect fluid solutions in which (i) the mass density is non-zero, (ii) the fluid four-velocity is expansion free, shear free, and geodesic, and (iii) the underlying manifold is simply connnected9 , namely the Einstein static universe (Hawking and Ellis [30]) and Gödel spacetime. (G¨ odel asserted this result, without proof, in [25]. Proofs can be found in Ozsváth [48] and 9 The

third condition is needed to rule out further examples that can be generated by identifying points.

178


Farnsworth and Kerr [19].) So G¨ odel spacetime itself is picked out if one adds the requirement that (iv) the rotation field of the fluid is non-vanishing. We next want to establish the existence of closed timelike curves in Gödel spacetime and characterize its timelike and null geodesics. To do so, it will be convenient to switch to a different coordinate representation of the metric. This one, cast in terms of a cylindrical coordinate system t˜, r, φ, z˜, makes manifest the rotational symmetry of G¨ odel spacetime about a particular axis, but hides its homogeneity: h i √ gab = 4 µ2 (da t˜)(db t˜) − (da r)(db r) − (da z˜)(db z˜) + (sh4 r − sh r2 )(da φ)(db φ) + 2 2 sh2 r (d(a t˜)(db) φ) .

(3.1.31)

(Here we write “ch” and “sh” for “cosh” and “sinh” respectively.) We have to be a bit careful here as to what we mean by a “coordinate system”. We are not quite talking about a 4-chart in the sense of section 1.1. Here is a more precise formulation. Let A be the “axis set” consisting of all points in M of the form Φ−1 (t, 0, 0, z), and let M − be the excised set M − A. We

claim that there exist smooth maps t˜: M → R

r : M − → R+

φ : M − → S1

z˜ : M → R

(3.1.32)

such that the composite map ∆ : M − → R × R+ × S 1 × R

(3.1.33)

determined by the rule q 7→ (t˜(q), r(q), φ(q), z˜(q)) is a diffeomorphism and (3.1.31) holds on M − . (Here

R+ is the set of reals that are strictly positive, and S 1 is identified, in the usual way, with R mod 2π.) Under these conditions, we can define coordinate vector fields (∂/∂ t˜)a , (∂/∂r)a , (∂/∂φ)a , (∂/∂ z˜)a much as we did in section 1.1.10 We shall use the following abbreviations for them: a a a a ∂ ∂ ∂ ∂ tã = φa = zã = . ra = ∂r ∂φ ∂ z˜ ∂ t˜ The radial coordinate r can be extended to a map r : M → R+ ∪ {0} that is, at least, continuous on the

axis A.

The relation between the new conditions and the old is given by the following conditions:

tan 10 So,

ex

=

y ex

=

z = φ t − 2 t˜ = + √ 2 2 2

ch 2r + (cos φ)(sh 2r), √ 2 (sin φ)(sh 2r), 2 z˜, e−2r tan

φ 2

where

(3.1.34) (3.1.35) t − 2 ˜t √ < π. 2 2 2

(3.1.36) (3.1.37)

for example, let q be any point in M − . Then s 7→ ∆−1 (t˜(q), r(q), φ(q) + s, z˜(q)) is a smooth curve through q. We

understand (∂/∂φ)a at q to be the tangent vector to the curve there.

179


With some work, one can show directly that these conditions do, in fact, properly define smooth maps over the domains indicated in (3.1.32)11 and use them to derive the expression for gab given in (3.1.31). (The details are worked out with great care in Stein [58].) We skip this work and make just two remarks about the conditions. Later, in an appendix, following Gödel [25], we shall establish the equivalence of the two coordinate representations a somewhat different way. It will involve a direct appeal to a coordinate-free description of the metric. First, it is clear from the first two conditions in the list why we need to restrict attention to M − . If x = y = 0, they will be satisfied iff r = 0. But if r = 0, those conditions impose no constraints on φ (and neither do the other conditions). So φ is not well defined on M − M − . (On the other hand, if either x 6= 0 or y 6= 0, then (3.1.34) and (3.1.35) determine unique values for both φ and r.)

Second, though the exact relation between t and t˜ is complex, their associated coordinate fields tã and ta are proportional to each other, i.e, we have tã = α ta for some α. This follows from the first three conditions. For when r, φ, z˜ are fixed, x, y, z are fixed as well. So every t˜-line (characterized by constant values of r, φ, z˜) is also a t-line (characterized by constant values for x, y, z). And it follows from (3.1.37) that the proportionality factor must be 2.12 So we have tã = 2 ta .

(3.1.38)

zã = 2 z a

(3.1.39)

We also have

from (3.1.36). Let us now accept as given the second coordinate representation of the Gödel metric (in terms of cylindrical coordinates). We shall work with it much as we did the first representation. Note that the inverse of the metric now comes out (in M − ) as # " √ 4 2 (sh r − sh r) 1 2 1 2 − t˜b t˜c − rb rc − z˜b z˜c − φb φc + t˜(b φc) . g bc = 4 µ2 (sh4 r + sh2 r) (sh4 r + sh2 r) (sh4 r + sh2 r) (3.1.40) Consider φa . Since φa φa = 4 µ2 (sh4 r − sh2 r),

(3.1.41)

it qualifies as spacelike, null, or timelike at a point q in M − depending on whether r(q) is less than, √ equal to, or greater than the critical value rc = ln(1 + 2) where sh assumes the value 1. The angular coordinate φ is defined only on M − , but we can smoothly extend φa itself to all of M by taking it to be the zero vector on M − M − , i.e., on the axis A. We shall understand it that way in what follows. 11 Strictly

speaking, the conditions define t˜ only on the restricted domain M − . But it can be smoothly extended to all of

M. follows from (3.1.37), specifically, that the difference (t − 2 t˜) is constant on every t˜-line, i.e., once r and φ are fixed, (t − 2 t˜) is fixed as well. So tñ ∇n (t − 2 t˜) = 0. It follows that α = α tn ∇n t = tñ ∇n t = tñ ∇n (2 t˜) = 2. 12 It

180


Where φa is timelike (and where it is null but non-zero) it qualifies as future-directed, because temporal √ orientation is determined by ta (or, equivalently, tã ), and tã φa = 4 2 µ2 sh2 r. (So tã φa > 0, unless r = 0.) tã and zã are, of course, Killing fields. We know that from before. So is φa . It is the generator of a φ

one-parameter family of (global) isometries {Γs }s∈S 1 defined by   ∆−1 (t˜(p), r(p), φ(p) + s, z˜(p)) if p ∈ M − , φ Γs (p) =  p if p ∈ A.

The three Killing fields under consideration have vanishing Lie brackets with one another: [tã , φa ] = [tã , zã ] = [φa , zã ] = 0.

(3.1.42)

(Once again, these relations follow most easily from proposition 3.1.1.) Now let p be any point on the φ axis A. The maps Γs all leave p fixed, and leave tã and zã fixed as well (by proposition 1.6.6). So if φ U is the two-dimensional subspace of Mp that is orthogonal to both tã and zã , the maps Γs induce a one-parameter family of rotations of U . And what is true here of p is true quite generally, because of homogeneity as formulated in (2). So we have the following isotropy claim. (5) G¨ odel spacetime is (globally) isotropic in the following sense: given any point p, and any two 1 2 unit spacelike vectors σ a and σ a at p that are orthogonal to both tã and z a , there is an isometry 1 2 ψ : M → M such that ψ(p) = p, ψ∗ (tã ) = tã , ψ∗ (˜ z a ) = zã , and ψ∗ (σ a ) = σ a . And now it is also clear, as announced, that Gödel spacetime admits closed timelike (and closed null) curves. Indeed, consider the set of (maximally extended) integral curves of φa . They are closed curves, characterized by constant values for t˜, r and z˜. We shall call them (or their images) G¨ odel circles. As we have just seen, they qualify as timelike if r > rc and null if r = rc . These particular curves are centered on the axis A. But by homogeneity, it follows that given any point in Gödel spacetime, there are closed timelike and closed null curves passing through the point. Indeed, we can make a much stronger assertion. The “causal structure” of G¨ odel spacetime is completely degenerate in the following sense. (6) Given any two points p and q in G¨ odel spacetime, there is a smooth, future-directed timelike curve that runs from p and q. (Hence, since we can always combine timelike curves that run in the two directions and smooth out the joints, there is a smooth closed timelike curve that contains p and q.) Thus a time traveler in G¨ odel spacetime can start at any point p, return to that point, and stop off at any other desired point q along the way. To see why (6) holds, consider figure 3.1.1. It gives, at least, a rough, qualitative picture of G¨ odel spacetime with one dimension suppressed. We may as well take the central line to be the axis A and take p to be a point on A. (By homogeneity once again, there is no loss in generality in doing so.) Notice first that given any other point p′ on A, no matter how “far down”,

181


matter worldline

p

p’

closed timelike line

Figure 3.1.1: G¨ odel spacetime with one dimension (the z˜ dimension) suppressed.

there is a smooth, future-directed timelike curve that runs from p to p′ . We can think of it as arising in three stages. (i) By moving “radially outward and upward” from p (i.e., along a future-directed timelike curve whose tangent vector field is of the form tã + α ra , with α positive13 ), we can reach a point p1 with coordinate value r > rc . At that radius, we know, φa is timelike and future-directed. So we can find an ǫ > 0 such that (−ǫ tã + φa ) is also timelike and future-directed there. (ii) Now consider the maximally extended, future-directed timelike curve γ through p1 whose tangent is everywhere equal to (−ǫ tã + φa ) (for that value of ǫ). It is a spiral-shaped curve of fixed radius, with “downward pitch”. By following γ far enough, we can teach a point p2 that is well “below” p′ . (We can overshoot as much as we might want.) Now, finally, (iii) we can reach p′ by working our way upward and inward from p2 via a curve whose tangent vector is the form tã + α ra , but now with α negative. It remains only to smooth out the “joints” at intermediate points p1 and p2 to arrive at a smooth timelike curve that, as required, runs from p to p′ . Now consider any point q. It might not be possible to reach q from p in the same simple way we went from p to p1 , i.e., along a future-directed timelike curve that moves radially outward and upward. p might be too “high” for that. But we can get around this problem by first moving to an intermediate point p′ on A sufficiently “far down” — we have established that that is possible — and then going from there to q. (This completes the argument for (6).) Other interesting features of G¨ odel spacetime are closely related to the existence of closed timelike curves. So, for example, a slice (in any relativistic spacetime) is a spacelike hypersurface that, as a subset of the background manifold, is closed. We can think of it as a candidate for a “global simultaneity slice”. It turns out that there are no slices in Gödel spacetime. More generally, given any relativistic spacetime, if it is temporally orientable and simply connected, and has smooth closed timelike curves through every point, then it does not admit any slices (Hawking and Ellis [30, p. 170]). 13 Note

that tã + α r a is timelike so long as α2 < 1.

182


Next we have the following basic fact. (7) There are no closed timelike or null geodesics in G¨ odel spacetime. We can easily confirm this, even before we characterize the class of timelike and null geodesics. It suffices (by homogeneity) to show that there are no closed timelike or closed null geodesics that pass through some particular point p on the axis A. Consider the set C = {q : r(q) < rc }. We shall call it the critical

cylinder surrounding A. We can establish our claim by showing two things: (i) all timelike geodesics that pass through p are fully contained within C, and all null geodesics that pass through p are fully contained within the closure of C; and (ii) there are no (non-trivial) closed causal curves within the closure of C. For (i), let γ be any timelike or null geodesic that passes through p, and let λa be its tangent field. We may as well assume that γ is future-directed (since otherwise we can run the argument on a new curve that results from reversing the orientation of γ). Since φa is a Killing field, the quantity λa φa is constant on γ. (Recall problem 1.9.6.) It is equal to 0 at p, since φa is the zero vector there. So it must be 0 everywhere. Now on the boundary of C (where r = rc ), φa is a non-zero, future-directed null vector. So its inner product there with any future-directed timelike vector is strictly positive. It follows that if γ is timelike, it can never reach the boundary of C. (If it did, we would have λa φa > 0 there.) It must stay within the (open) set C. Similarly, at all points outside the closure of C, φa is a future-directed timelike vector. So its inner product with all future-directed causal vectors (even null ones) is strictly positive. And therefore, if γ is null, it must remain within the closure C. (As we shall see in a moment, null geodesics through p do periodically intersect the boundary of C.) For (ii), note that, by (3.1.40), g ab (∇a t˜) (∇b t˜) = −

1 (sh4 r − sh2 r) . 4 µ2 (sh4 r + sh2 r)

So ∇a t˜ is timelike within C and null (and non-zero) on the boundary of the set. It is future-directed both in C and on its boundary (since tã ∇a t˜ = 1). Now let γ be any non-trivial future-directed causal curve that passes through p, and let λa be its tangent field. Then (since λa and ∇a t˜ are co-oriented), we have λn ∇n t˜ > 0 at all points in C and λn ∇n t˜ ≥ 0 at all points on the boundary of the set. So γ cannot possibly stay within the closure of C and still close back on itself.

Now, finally, let us characterize the set of all timelike and null geodesics in Gödel spacetime. The zã direction is not very interesting here, and we may as well restrict attention to curves that fall within a zã = constant submanifold, i.e., curves whose tangent fields are orthogonal to zã (or equivalently to z a ).14 14 Given

any smooth curve s 7→ Φ−1 (t(s), x(s), y(s), z(s)) in G¨ odel spacetime, it qualifies as a geodesic iff (i) z(s) is of

the form z(s) = z0 + k s, for some numbers z0 and k, and (ii) the projected curve s 7→ Φ−1 (t(s), x(s), y(s), z0 ) qualifies as

a geodesic. This follows because ∇a z b = 0.

183


We shall first consider certain examples that admit a particularly simple description. Then we shall argue that they are, up to isometry (and reparametrization), the only ones. A small bit of computation is involved. For that we need the following simple results that are the counterparts to ones presented earlier for the first set of coordinates. At points in M − , where r > 0, we have φb ∇a φb φa ∇a φb φa ∇a t˜b

= 4 µ2 [(sh4 r − sh2 r)∇b φ +

√ 2 sh2 r ∇b t˜],

(3.1.43)

√ = 4 µ2 [ (4 sh3 r − 2 sh r)(ch r)(∇[a r)(∇b] φ) + 2 2 (sh r)(ch r) (∇[a r)(∇b] t˜) ], = −4 µ2 (2 sh3 r − sh r)(ch r) ∇b r, √ = tã ∇a φb = −4 2 µ2 (sh r)(ch r) ∇b r.

(3.1.44) (3.1.45) (3.1.46)

(For the second equation, we use the fact that φa is a Killing field and, so, ∇(a φb) = 0. For the fourth, we use (3.1.42).)

Consider fields of the form tã + k φa , where k is some real number. Their integral curves are “helices” on which r and z˜ are constant (since tã ∇a r = tã ∇a z˜ = 0, and similarly for φa ). Our goal is to show that some of these helices — characterized by particular choices for k and r — are causal geodesics. Let k and r be fixed, and let γ be an integral curve of tã + k φa associated with these values. Then, we have (tã + k φa )(tã + k φa ) = 4 µ2 [ k 2 (sh4 r − sh2 r) + 2

√ 2 (sh2 r) k + 1 ]

(3.1.47)

and (by (3.1.45), (3.1.46), and the fact that tã is a geodesic field), √ 2 k [ −4 2 µ2 (sh r)(ch r) ∇b r ]

(tã + k φa )∇a (t˜b + k φb ) =

+ k 2 [ −4 µ2 (2 sh3 r − sh r)(ch r) ∇b r ] √ (3.1.48) −4 µ2 k (sh r)(ch r) [2 2 + k (2 sh2 r − 1)] ∇b r.

=

Thus γ is a geodesic iff k = 0 (in which case it is just an integral curve of tã ), or r = 0 (in which case, again, it is an integral curve of tã , now on the axis), or √ k (2 sh2 r − 1) + 2 2 = 0.

(3.1.49)

It is a null geodesic iff this condition holds and the right side of (3.1.47) is 0. That leaves us with two equations in two unknowns. They yield γ is a null geodesic

⇐⇒

√ ( 2 − 1) sh r = 2 2

and

k = 2(1 +

√

2)

or, equivalently (since sh 2r = 2 (sh r)(ch r)), γ is a null geodesic

⇐⇒

r =

rc 2

and

k = 2(1 +

√ 2).

Similarly, after excluding the trivial cases where k = 0 or r = 0, , we have γ is a timelike geodesic

⇐⇒

r
(tã φa )2 ≥ 0, since tã is timelike and φa is spacelike on R. So l1 > 0 and 23 Our

proof proceeds by way of a low-brow calculation. For a more insightful argument, see Ashtekar and Magnon [3].

198


l2 < 0. Moreover, l2 < k < l1 . (Consider the quadratic function f (x) = (tã + x φa )(tã + x φa ). It is concave downward because (φa φa ) is negative. So, since f (k) > 0 and f (l1 ) = f (l2 ) = 0, it must be the case that k falls between l1 and l2 .) So (l1 − k) > 0

(l2 − k) < 0.

and

It follows from our initial assumptions about the background spacetime (M, gab ) that there there exist smooth coordinate maps t˜ : R → R and φ : R → R (mod 2π) on the orbit cylinder R such that tã ∇a t˜ = φa ∇a φ = 1 and tã ∇a φ = φa ∇a t˜ = 0.24 Now consider the hybrid field φ′ : R → R (mod 2π) defined by

φ′ = (φ − k t˜) (mod 2π). It is adapted to (R, k) in the sense that it is constant on all integral curves of (tã + k φa ): (tñ + k φn )∇n (φ − k t˜) = tñ ∇n (−k t˜) + (k φn )∇n φ = 0. In particular, φ′ is constant on γ. In contrast, φ′ increases (resp. decreases) uniformly with respect to elapsed parameter distance along λ1 (resp. λ2 ) since (tñ + li φn )∇n φ′ = (li − k). (It follows, as claimed above, that λ1 and λ2 must reintersect γ.)

Let the points p0 , p1 , and p2 have respective t˜, φ′ coordinates (t˜0 , φ′ ), (t˜1 , φ′ ), and (t˜2 , φ′ ). They share a common φ′ coordinate since φ′ is constant on γ. But φ′ increases along λ1 in the stretch between p0 and p1 — it goes from 0 to 2π. Similarly, φ′ decreases along λ2 in the stretch between p0 and p2 — it goes from 0 to −2π. The coordinate t˜ increases along all three curves, γ, λ1 , and λ2 . (Indeed, we have (tñ + k φn )∇n t˜ = (tñ + li φn )∇n t˜ = 1.) So we can think of the curves as parametrized by t˜ and consider the rate of change of φ′ with respect to t˜ on them. This rate of change on λi is (by the chain rule) dφ′ (tñ + li φn )∇n (φ − k t˜) = = (li − k). dt˜ (tñ + li φn )∇n t˜ So, considering the total change of φ′ along λ1 and λ2 , we have ′

2π

dφ = (t˜1 − t˜0 ) dt˜

−2π

dφ = (t˜2 − t˜0 ) dt˜

|on λ1

= (t˜1 − t˜0 ) (l1 − k),

|on λ2

= (t˜2 − t˜0 ) (l2 − k).

′

24 We

can introduce the coordinates as follows. Pick any initial point on R and take its coordinates to be t˜ = 0, φ = 0.

Given any other point on R, we can “get to it” from the initial point by moving a certain (signed) parameter distance along an integral curve of tã and moving a certain (signed) parameter distance along an integral curve of φa . It does not matter in what order we perform the operations because the fields tã and φa have a vanishing Lie bracket. We take the respective parameter distances to be the t˜ and φ coordinates of the new point.

199


It follows that

2 π (l1 + l2 − 2k) t˜1 − t˜2 = . (l1 − k) (l2 − k)

Hence, by (3.2.6) and (3.2.7), p1 = p2

⇐⇒

t˜1 = t˜2

⇐⇒

(l1 + l2 − 2k) = 0

⇐⇒

k=−

(tã φa ) . (φn φn )

This gives us (3.2.5). Now we consider the two criteria in the special case of Gödel spacetime. We start with a calculation. Proposition 3.2.3. Let ǫabcd be a volume element on G¨ odel spacetime and let η a be the field tã + k φc for some choice of k. Then h √ √ i ǫabcd η[b ∇c ηd] = ± 2 k 2 2 sh4 r + k (2 sh2 r − 1) + 2 zã

(3.2.8)

where, as in the previous section, zã = (∂ /∂ z˜)a .

Note that in the special case where k = 0, this yields √ ǫabcd t˜[b ∇c t˜d] = ± 2 2 zã . If we re-express this in terms of tâ = ta /µ = tã /(2µ) and z a = zã /2, and choose a volume element so that the right side sign is +1, we recover (3.1.30), i.e., 1 1 abcd ˆ z a. ǫ tb ∇c tˆd = √ 2 2 µ2 Proof. As before, let A be the set of axis points in Gödel spacetime where r = 0, and let M − be the complement set M − A. The vector fields tã = (∂ /∂ t˜)a

ra = (∂ /∂ r)a

φa = (∂ /∂ φ)a

zã = (∂ /∂ z˜)a

are linearly independent on M − . So we can express ǫabcd in the form ǫabcd = f t˜[a rb φc z˜d] on M − . We can determine f , up to sign, as follows. We certainly have −(4!) = ǫabcd ǫabcd = f 2 t˜[a rb φc z˜d] t˜[a rb φc z˜d] = f 2 t˜[a rb φc z˜d] tã rb φc z˜d . And by (3.1.31), tã

=

√ 4 µ2 [ 2 sh2 r ∇a φ + ∇a t˜],

rb

=

4 µ2 ∇b r,

φc

=

4 µ2 [(sh4 r − sh2 r)∇c φ +

z˜d

=

4 µ2 ∇d z˜.

√

2 sh2 r ∇c t˜],

200


So −(4!) = =

f 2 t˜[a rb φc z˜d] (4 µ2 )4 (sh4 r − sh2 r) − 2 sh4 r (∇a t˜) (∇b r) (∇c φ) (∇d z˜) 1 1 − f 2 (4 µ2 )4 (sh4 r + sh2 r) = − f 2 (4 µ2 )4 (sh2 r) (ch2 r) . 4! 4!

Thus, on M − , we have ǫabcd = ±

4! t˜[a rb φc z˜d] . 16 µ4 (sh r)(ch r)

(3.2.9)

Next, we derive an expression for η[b ∇c ηd] = t˜[a ∇b t˜c] + k t˜[a ∇b φc] + k φ[a ∇b t˜c] + k 2 φ[a ∇b φc]

(3.2.10)

on M − . Note first that ∇b φc

=

∇b t˜c

=

√ 4 µ2 [ (4 sh3 r − 2 sh r)(ch r)(∇[b r)(∇c] φ) + 2 2 (sh r)(ch r) (∇[b r)(∇c] t˜) ], √ 4 µ2 [ 2 2 (sh r)(ch r)(∇[b r)(∇c] φ) ],

both hold on M − . (The first is (3.1.44). The second is derived similarly, using the fact that ∇(b t˜c) = 0.) These expressions, together with the preceding ones for tã and φa , yield t˜[a ∇b t˜c] t˜[a ∇b φc] φ[a ∇b t˜c] φ[a ∇b φc]

= 16 µ4 2

√ 2 (sh r)(ch r) (∇[a t˜)(∇b r)(∇c] φ),

4

3

3

= 16 µ [ (4 sh r − 2 sh r)(ch r) − 4 (sh r)(ch r)] (∇[a t˜)(∇b r)(∇c] φ), 4

3

= 16 µ 4 (sh r)(ch r) (∇[a t˜)(∇b r)(∇c] φ), √ = 16 µ4 [ 2 (sh2 r)(ch r)(4 sh3 r − 2 sh r) √ − 2 2 (sh r)(ch r)(4 sh3 r − 2 sh r) ] (∇[a t˜)(∇b r)(∇c] φ).

If we insert these expressions in (3.2.10), we arrive at: h √ √ i η[b ∇c ηd] = 32 µ4 (sh r)(ch r) k 2 2 sh4 r + k (2 sh2 r − 1) + 2 (∇[a t˜)(∇b r)(∇c] φ).

(3.2.11) (3.2.12) (3.2.13)

(3.2.14)

(3.2.15)

Finally, combining this result with (3.2.9) yields ǫabcd η[b ∇c ηd]

= =

4! t˜[a rb φc z˜d] η[b ∇c ηd] 16 µ4 (sh r)(ch r) √ √ ± 2 (k 2 2 sh4 r + k (2 sh2 r − 1) + 2 ) zã

±

(3.2.16)

on M − . Since both η a and the final vector field in (3.2.16) are smooth (everywhere), the equation must hold on A as well. Our desired characterization result for Gödel spacetime follows as a corollary. (For clause (2), recall that sh2 rc = 1.) Proposition 3.2.4. Let R be a striated orbit cylinder in G¨ odel spacetime generated by tã and φa . It is characterized by particular values for r (where 0 < r < rc ) and z˜. Let k be such that η a = tã + k φa is timelike on R. Then the following both hold.

201


(1) (R, k) is non-rotating according to the ZAM criterion (2) (R, k) is non-rotating according to the CIR criterion rc r < 2

and

⇐⇒

k =

√ 2 . (1 − sh2 r)

⇐⇒

√ 2 p k = . 2 (1 − 2 sh r) + 1 − sh2 (2r) 2

Proof. Note that our assumption that tã + k φa is timelike on R comes out as the assumption that the

relation

√ k 2 (sh4 r − sh2 r) + k 2 2 sh2 r + 1 > 0

(3.2.17)

holds there. (We are making use of (3.1.31) here and shall do so repeatedly in what follows.) (R, k) qualifies as non-rotating according to the ZAM criterion iff (tã + k φa )φa = 0 on R. The

latter condition comes out as

√

2 sh2 r + k (sh4 r − sh2 r) = 0. √ Moreover, as is easy to check, if k = 2/(1 − sh2 r), then (3.2.17) is automatically satisfied, i.e., (3.2.17)

imposes no further constraint on k in this case. So we have clause (1).

ǫ

Next, (R, k) is non-rotating according to the CIR criterion iff η[b ∇c ηd] = 0 on R or, equivalently, if

abcd

η[b ∇c ηd] vanishes there (for either choice of ǫabcd ). We know from the preceding proposition that

this is the case iff

k2

√ √ 2 sh4 r + k (2 sh2 r − 1) + 2 = 0

on R. This equation has two roots: p (1 − 2 sh2 r) − 1 − sh2 (2r) √ k1 = 2 2 sh4 r

and

(3.2.18)

p (1 − 2 sh2 r) + 1 − sh2 (2r) √ k2 = . 2 2 sh4 r

So it is a necessary condition for (R, k) to be non-rotating according to the CIR criterion (for any choice of k) that sh2 2r ≤ 1 or, equivalently, that r ≤ rc /2. So assume this condition holds. We claim that the

root k2 can be ruled out because it leads to a violation of (3.2.17). We also claim that k1 is compatible with that inequality if we further restrict r so that sh2 2r < 1. To see this, note that in the presence of (3.2.18), (3.2.17) holds iff k2

√ 2 sh2 r − k (2 sh2 r + 1) < 0

and this holds, in turn, iff

2 sh2 r + 1 √ . (3.2.19) 2 sh2 r With a bit of straightforward algebra, one can easily check that k2 violates this inequality but that k1 √ satisfies it if sh2 2r < 1. Finally, note that X = [(1 − 2 sh2 r) + 1 − sh2 2r ] 6= 0. So we have √ 2 2 X p = k1 = k1 . X (1 − 2 sh2 r) + 1 − sh2 (2r) 0 < k
2 M . If we think of Kerr spacetime as representing the spacetime structure surrounding a rotating black hole, our interest will be in small rings that are positioned close to the axis of rotational symmetry (where sin2 θ is small) and far away from the center (where r is large). There we can sidestep all complexities having to do with horizons and singularities. The proposition we need is the following. Proposition 3.3.3. Let p be an axis point in Kerr spacetime with coordinates t˜ and r > 2M . Then kcrit (p) =

2M ra . (r2 + a2 )2

(3.3.2)

(So kcrit does not assume the same value at all axis points.) Proof. We can certainly verify (3.3.2) directly by computing (∇b t˜c )(∇b φc ) (∇m φn )(∇m φn ) at p and then invoking (3.3.1). But we can save ourselves a bit of work with an alternate approach that focuses attention on the smooth function f : M − → R defined by f = −

2M ra (tã φa ) = 2 . n 2 2 (φ φn ) (r + a )ρ + 2 M r a2 sin2 θ

Consider the field η a = tã + f φa on M − . We claim that it can be expressed in the form ηa =

∇a t˜ . (∇n t˜)(∇n t˜)

(3.3.3)

212


To see this, let D = (tã tã )(φa φa ) − (tñ φn )2 . Clearly, D < 0 on M − (since φa is spacelike there). We have

∇a t˜ =

1 n (φ φn ) tã − (tñ φn ) φa . D

(This follows since both sides yield the same result when contracted with tã , φa , ra , and θa .) Hence (∇n t˜)(∇n t˜) = and, therefore,

1 n (φ φn ) D

∇a t˜ (φn φn ) tã − (tñ φn ) φa = = ηa . n (φn φn ) (∇n t˜)(∇ t˜)

as claimed. The right side of (3.3.3) has the form g ∇a t˜. It follows that η[a ∇b ηc] = 0 everywhere on M − . Now f and η a can be smoothly extended to A. At p, the extended function assumes the value f (p) =

2M ra (r2 + a2 )2

(since, once again, axis points here are ones where sin θ = 0). So, at p, the extended vector field satisfies 0 = η[a ∇b ηc] = t˜[a ∇b t˜c] + f (p) t˜[a ∇b φc] + f (p) φ[a ∇b t˜c] + f (p)2 φ[a ∇b φc] . But we know from proposition 3.3.1 that the final expression on the right can be 0 only if f (p) = kcrit (p). So we are done. Our main result now follows as an immediate corollary. Proposition 3.3.4. (No-Go Theorem) There is no criterion of ring non-rotation on Kerr spacetime that satisfies the relative rotation, limit, and non vacuity conditions. It is intended to bear this interpretation: given any (non-vacuous) generalized criterion of ring nonrotation in Kerr spacetime, to the extent that it gives “correct” attributions of non-rotation in the limit for “infinitesimally small” rings — the domain where one does have an unambiguous notion of non-rotation — it must violate the relative rotation condition.

Appendix: The Proof of Proposition 3.3.1. Here we prove proposition 3.3.1. It will be convenient to collect a few facts first that will be used in the proof. Proposition 3.3.5. Let (M, gab ) be a stationary, axi-symmetric spacetime with Killing fields tã and φa . Let p be an axis point. (So φa = 0 at p.) Let ǫabcd be a volume element defined on some open set O containing p, and let σ a be the smooth field on O defined by σ a = ǫabcd t˜b ∇c φd . Then at p, (1) σ a 6= 0

213


(2) ∇a φb =

1 ǫabcd t˜c σ d . ˜ 2 (tn tñ )

Furthermore, given any smooth field ψ a (defined on some open set containing p), if £φ ψ a = 0 at p, then it must be of the form ψ a = k1 tã + k2 σ a at p. Proof. Note that σ a is orthogonal to tã and φa throughout O. (The first claim follows just because ǫabcd is anti-symmetric, and the second by clause (vi) in our characterization of stationary, axi-symmetric spacetimes.) Note, as well, that

1 t˜[a ∇b φc] = ǫabcd σ d 6

(3.3.4)

throughout O. (We get this by contracting both sides of σ d = ǫdmnp t˜m ∇n φp with ǫabcd .) Now we argue for (1). Suppose that σ a = 0 at p. Then, by (3.3.4),

1 ã ˜ 0 = tã t˜[a ∇b φc] = (t ta )∇b φc + t˜c tã ∇a φb − t˜b tã ∇a φc 3

at p. Now tã ∇a φb = φa ∇a t˜b everywhere on O (since the fields tã and φa have a vanishing Lie bracket),

and φa = 0 at p. So the second and third terms on the right vanish there. Thus ∇a φb = 0 at p. But this

is impossible. For given any Killing field κa on the (connected) manifold M , if κa and ∇a κb both vanish

at any one point, then they must vanish everywhere. (See Wald [60, p. 443].) And that is not possible

in the present case because φa is spacelike at all non-axis points (and there exist some non axis points). So we have (1). And for (2) we need only contract both sides of (3.3.4) with t˜c , expand the left side, and use much the same argument we have just used to show that two terms in the expansion vanish. Finally, let ψ a be a smooth field (defined on some open set containing p) such that £φ ψ a = 0 at p. Then ψ a ∇a φb = φa ∇a ψb = 0 at p (since, once again, φa = 0 at p). Hence, by (2), ǫabcd ψ a t˜c σ d = 0. So

the three vectors ψ a , tã , and σ a are linearly dependent at p. Since tã and σ a are non-zero at p, ψ a can be expressed as a linear combination of them at p. Now for the proof of proposition 3.3.1. The formulation, once again, is as follows. Let (M, gab ) be a stationary, axi-symmetric spacetime with Killing fields tã and φa . Further, let p be a point at which φa = 0. Then there is a unique number k such that η a = tã + k φa is non-rotating (η[a ∇b ηc] = 0) at p, and that number is −

(∇b t˜c )(∇b φc ) . (∇m φn )(∇m φn )

Proof. For the first claim, what we need to show is that there a unique k such that t˜[a ∇b t˜c] + k t˜[a ∇b φc] = 0

(3.3.5)

at p. (This is equivalent to η[a ∇b ηc] = 0 at p since φa = 0 there.) We know from clause (1) of the preceding proposition and (3.3.4) that t˜[a ∇b φc] 6= 0 at p. So uniqueness is immediate. For existence,

214


let ǫabcd be a volume element defined on some open set containing p, let σ a = ǫabcd t˜b ∇c φd (as in the preceding proposition), and let ω a = ǫabcd t˜b ∇c t˜d . The new field ω a is orthogonal to tã . And it is Lie derived by φa , i.e., £φ ω a = 0 (since φa is a Killing field that commutes with tã ). So, by the preceding proposition, there is a number k2 such that ǫabcd t˜b ∇c t˜d = ω a = k2 ǫabcd t˜b ∇c φd or, equivalently, t˜[a ∇b t˜c] = k2 t˜[a ∇b φc] at p. Thus (3.3.5) holds at p iff k = −k2 . Now we compute k2 . Contracting the preceding line with tã ∇b φc , and then dividing by (tã tã ), yields (∇b t˜c )(∇b φc ) = k2 (∇b φc )(∇b φc ) at p. So, to complete the proof, we need only verify that (∇b φc )(∇b φc ) 6= 0 at p. But this follows from

the preceding proposition. By clause (2) we have (∇b φc )(∇b φc ) =

1 1 ǫbcmn t˜m σ n (∇b φc ) = − (σn σ n ) 2 (tñ tñ ) 2 (tñ tñ )

at p. And σ a is spacelike at p, since it is orthogonal to tã and (by clause (1)) non-zero there.

Chapter 4

Newtonian Gravitation Theory The “geometrized” formulation of Newtonian gravitation theory — also known as “Newton-Cartan theory” — was first introduced by Cartan [5, 6] and Friedrichs [21] and later developed by Dautcourt [10], Dixon [11], Dombrowski and Horneffer [13], Ehlers [15], Havas [28], K¨ unzle [34, 35], Lottermoser [37], Trautman [59], and others. It is significant for several reasons. First, it shows that several features of relativity theory once thought to be uniquely characteristic of it do not distinguish it from (a suitably reformulated version of) Newtonian gravitation theory. The latter too can be cast as a “generally covariant” theory in which (i) gravity emerges as a manifestation of spacetime curvature, and (ii) spacetime structure is “dynamical” in the sensed that it participates in the unfolding of physics rather than being a fixed backdrop against which it unfolds. Second, it clarifies the gauge status of the Newtonian gravitational potential. In the geometrized g

formulation of Newtonian theory, one works with a single curved derivative operator ∇. It can be de-

composed (in a sense) into two pieces — a flat derivative operator ∇ and a gravitational potential φ — to recover the standard formulation of the theory. But in the absence of special boundary conditions, the

decomposition will not be unique. Physically, there is no unique way to divide into “inertial” and “gravitational” components the forces experienced by particles. Neither has any direct physical significance. Only their “sum” does. It is an attractive feature of the geometrized formulation that it trades in two gauge quantities for this sum. (See the discussion at the end of section 4.2.) Third, the clarification just described also leads to a solution, or dissolution, of an old problem about Newtonian gravitation theory, namely the apparent breakdown of the theory when applied (in cosmology) to a hypothetically infinite, homogeneous mass distribution. (See section 4.4.) Fourth, it allows one to make precise in coordinate-free, geometric language the standard claim that Newtonian gravitation theory (or, at least, a certain generalized version of it) is the “classical limit” of general relativity. (See K¨ unzle [35], Ehlers [15], and Lottermoser [37].)

215

216

CHAPTER 4. NEWTONIAN GRAVITATION THEORY

4.1

Classical Spacetimes

We begin our discussion by characterizing a new class of geometric models for the spacetime structure of our universe (or subregions thereof) that is broad enough to include the models considered in both the standard and geometrized versions of Newtonian gravitation theory. We take a classical spacetime to be a structure (M, tab , hab , ∇) where (i) M is a smooth, connected,

four-dimensional manifold, (ii) tab is a smooth, symmetric field on M of signature (1, 0, 0, 0), (iii) hab

is a smooth, symmetric field on M of signature (0, 1, 1, 1), (iv) ∇ is a derivative operator on M , and (v) the following two conditions hold:

hab tbc ∇a tbc = 0

=

0

(4.1.1) ∇a hbc = 0.

and

(4.1.2)

We refer to them, respectively, as the “orthogonality” and “compatibility” conditions. M is interpreted as the manifold of point events (as before). Collectively, the objects tab , hab , and ∇

on M represent the spacetime structure presupposed by classical Galilean relativistic dynamics. It will soon emerge how they do so. We need to explain what we mean by the “signatures” of tab and hab , since we are using the term here in a new, somewhat extended sense. The signature condition for tab is the requirement that, at 1

4

every point in M , the tangent space there have a basis ξ a , ..., ξ a such that, for all i and j in {1, 2, 3, 4}, i

j

tab ξ a ξ b = 0 if i 6= j, and

  1 tab ξ a ξ b =  0 i

if i = 1

i

if i = 2, 3, 4.

(We shall call this an “orthonormal basis” for tab , though this does involve a slight extension of ordinary P4 i i P4 i i usage.) Hence, given any vectors µa = i=1 µ ξ a and ν a = i=1 ν ξ a at the point, 1 1

tab µa ν b = µ ν

(4.1.3)

and 1

tab µa µb = (µ)2 ≥ 0.

(4.1.4)

Notice that tab is not a metric as defined in section 1.9, since it does not satisfy the required non1

2

4

degeneracy condition. (For example, if the vectors ξ a , ..., ξ a are as above at some point, then tab ξ a = 0 2

there, even though ξ a 6= 0.) The signature condition for hab , similarly, is the requirement that, at every point, the cotangent space 1

4

i

j

there have a basis σ a , ..., σ a such that, for all i and j in {1, 2, 3, 4}, hab σ a σ b = 0 if i 6= j, and   0 if i = 1 i i hab σ a σ b =  1 if i = 2, 3, 4.

217


(We shall extend ordinary usage once again and call this an “orthonormal basis” for hab .) Hence, given P4 i i P4 i i any vectors αa = i=1 α σ a and β a = i=1 β σ a at the point, 2 2

3 3

4 4

hab αa βb = α β + α β + α β

and

2

3

(4.1.5)

4

hab αa αb = (α)2 + (α)2 + (α)2 ≥ 0.

(4.1.6)

Notice, too, that hab is not the inverse of a metric (in the sense of section 1.9), i.e., there is no field hab 1

4

1

such that hab hbc = δ c a . (Why? If σa , ..., σ a are as above at some point, then hab σ a = 0. Hence, if there 1

1

1

were a tensor hab at the point such that hab hbc = δ c a , it would follow that 0 = hab hbc σ c = δ c a σ c = σ a , 1

4

contradicting the assumption that σ a , ..., σ a form a basis of the cotangent space there.) In what follows, let (M, tab , hab , ∇) be a fixed classical spacetime. Consider, first, tab . We can think of it as a “temporal metric”, even though it is not a metric in the 1

sense of section 1.9. Given any vector ξ a at a point, we take its “temporal length” to be (tab ξ a ξ b ) 2 . (We know from (4.1.4) that (tab ξ a ξ b ) must be non-negative.) We further classify ξ a as either timelike or spacelike depending on whether its temporal length is positive or zero. It follows from the signature of tab 1

4

that the subspace of spacelike vectors at any point is three-dimensional. (For if ξa , ..., ξa is an orthonormal 1

basis for tab there, ξa is timelike, and the remaining three are spacelike.) Notice too that at any point we 1

4

can find a co-vector ta , unique up to sign, such that tab = ta tb . (Again, let ξ a , ..., ξ a be an orthonormal 1

basis for tab at the point. Then ta = ± tan ξ n satisfies the stated condition. Conversely, if tab = ta tb , then 1

1

1

1

1

1

contraction with ξ a ξ b yields 1 = (ta ξ a )2 . So ta ξ a = ±1 and, hence, tab ξ b = ta (tb ξ b ) = ± ta .)

So far we have considered only the decomposition tab = ta tb at individual points of M . We say that (M, tab , hab , ∇) is temporally orientable if there exists a continuous (globally defined) vector field ta that

satisfies the decomposition condition at every point. (Our assumptions to this point do not guarantee existence.) Any such field ta (which must, in fact, be smooth since tab is) will be called a temporal orientation. A timelike vector ξ a qualifies as future-directed relative to ta if ta ξ a > 0; otherwise it is past-directed. If a classical spacetime admits one temporal orientation ta , then it admits two altogether, namely ta and −ta . In what follows, we shall restrict attention to classical spacetimes that are temporally orientable and in which a temporal orientation has been selected. (We shall say, for example, “consider the classical spacetime (M, ta , hab , ∇) ...”.) The orthogonality condition and the first compatibility condition can then

be formulated directly in terms of ta :

hab tb

=

0,

(4.1.7)

∇a tb

=

0.

(4.1.8)

(These follow easily from the original formulations.)


218

Clearly, we understand a smooth curve to be timelike (respectively spacelike) if its tangent vectors are of this character at every point. And a timelike curve is understood to be future-directed (respectively past-directed ) if its tangent vectors are so at every point. From the compatibility condition, it follows that ta is closed, i.e. ∇[a tb] = 0. So (by proposition

1.8.3), at least locally, it must be exact, i.e., of the form ta = ∇a t for some smooth function t. We call

any such function a time function. Any two time functions t and t′ defined on a (common) open set can

differ only by a constant, i.e., there must be a number k such that t′ (p) = t(p) + k for all p in the set. Given any time function t, and any smooth, future-directed timelike curve γ : [s1 , s2 ] → M with tangent

field ξ a (whose image falls within the domain of t), the temporal length of γ is given by Z s2 Z s2 Z s2 d(t ◦ γ) (ξ a ∇a t) ds = (ta ξ a ) ds = ds = t(γ(s2 )) − t(γ(s1 )), ds s1 s1 s1

i.e., it depends only on the endpoints of the curve. This shows that, at least locally, we have a well-defined, path-independent notion of “temporal distance” between points. Let us say that a hypersurface S in M is spacelike if, at all points of S, all vectors tangent to S are spacelike. Notice that the defining condition is equivalent to the requirement that all time functions be constant on S. (A time function t is constant on S iff, given any vector ξ a tangent to S at some point of S, ξ a ∇a t = 0. But ta ξ a = ξ a ∇a t. So the latter condition holds iff all vectors tangent to S are spacelike.)

We can think of spacelike hypersurfaces as (at least local) “simultaneity slices”.

If M is simply connected, then there must exist a globally defined time function t : M → R. In this

case, spacetime can be decomposed into a one-parameter family of global (t = constant) simultaneity slices. One can speak of “space” at a given “time”. A different choice of (globally defined) time function would result in a different zero-point for the time scale, but would induce the same simultaneity slices and the same temporal distances between points on them. We are now in a position to formulate interpretive principles corresponding to (C1), (P1), and (P2). (Recall our discussion in sections 2.1 and 2.3.) For all smooth curves γ : I → M , (C1′ ) γ is timelike iff its image γ[I] could be the worldline of a point particle. (P1′ ) γ can be reparametrized so as to be a timelike geodesic (with respect to ∇) iff γ[I] could be the worldline of a free point particle.1

(P2′ ) Clocks record the tab -length of their worldlines. Two points should be noted. First, in (C1′ ) and (P1′ ), we make reference to “point particles” without qualification, whereas previously we needed to restrict attention to particles with mass m > 0. Here there 1 We

have seen (proposition 2.5.2) that it is possible, in a sense, to recover principle (P1) as a theorem in general relativity.

Similarly, one can recover (P1′ ) as a theorem in geometrized Newtonian gravitation theory. Indeed, one can prove a result that is a close counterpart to proposition 2.5.2 (Weatherall [61]).

219


are no zero mass particles to consider, and no null curves whose images might serve as their worldlines. Second, there is an ambiguity as to what we mean by a “free” particle in (P1′ ). In the standard formulation of Newtonian gravitation theory, particles subject to a (non-vanishing) gravitational force do not count as free. But on the geometrized formulation, as in relativity theory, they do. In what follows, unless indication is given to the contrary, we shall understand a “timelike curve” to be smooth, future-directed, and parametrized by its tab –length. In this case, its tangent field ξ a satisfies the normalization condition ta ξ a = 1. And in this case, if a particle happens to have the image of the curve as its worldline, then we call ξ a the four-velocity field of the particle, and call ξ n ∇n ξ a its four-acceleration

field. If the particle has mass m, then its four-acceleration field satisfies the equation of motion F a = m ξ n ∇n ξ a ,

(4.1.9)

where F a is a spacelike vector field (on the image of its worldline) that represents the net force acting on the particle. This is our version of Newton’s second law of motion. (Recall (2.4.13).) Note that the equation makes geometric sense because the four-acceleration field is spacelike. (For, by the first compatibility condition, ta ξ n ∇n ξ a = ξ n ∇n (ta ξ a ) = ξ n ∇n (1) = 0.) Now consider hab . It serves as a spatial metric, but just how it does so is a bit tricky. In Galilean relativistic mechanics, we have no notion of spatial length for timelike vectors, e.g., four-velocity vectors, since having one is tantamount to a notion of absolute rest. (We can take a particle to be “at rest” if its four-velocity field has spatial length 0 everywhere.) But we do have a notion of spatial length for spacelike vectors, e.g., four-acceleration vectors. (We can, for example, use measuring rods to determine distances between simultaneous events.) The field hab gives us one without the other. 1

We cannot take the spatial length of a vector µa to be (hab µa µb ) 2 because the latter is not well defined. (As we have seen, there does not exist a field hab satisfying hab hbc = δ ac .) But if µa is spacelike, we can use hab to assign a spatial length to it indirectly. Here we need a small result about spacelike vectors. Proposition 4.1.1. Let (M, ta , hab , ∇) be a classical spacetime. Then the following conditions hold at all points in M .

(1) For all σb , hab σb = 0 iff σb is a multiple of tb . (2) For all µa , µa is spacelike iff there is a σb such that hab σb = µa . (3) For all σb and σ ′b , if hab σb = hab σ ′b , then hab σa σb = hab σ ′a σ ′b . Proof. The “if” halves of (1) and (2) follow immediately from the orthogonality condition (4.1.7). For 1

4

the “only if” half of (1), let σ a , ..., σ a be an orthonormal basis for hab in the sense discussed above. (So j

i

1

1

i

1

i

hab σ a σ b = 0 if i 6= j, hab σ a σ b = 0, and hab σ a σ b = 1 for i = 2, 3, 4.) We can take σ a to be ta , since the 1 1

2

2

3

3

4

4

latter satisfies the required conditions. Now consider any vector σb = k tb + k σ b + k σ b + k σ b , and assume 2

2

3

3

4

4

hab σb = 0. Then, by the orthogonality condition, k (hab σ b ) + k (hab σ b ) + k (hab σ b ) = 0. Contraction i

i

1 1

with σ b yields k = 0 for i = 2, 3, 4. So σb = k tb .

220


The “only if” half of (2) follows by dimensionality considerations. At any point in M , we can construe h

ab

as a linear map from the cotangent space Vb there to the tangent space V a . Every vector in the image 2

3

4

space hab [Vb ] is spacelike (by the “if” half of (2)). Moreover, hab [Vb ] is three-dimensional. (If σ b , σ b , σ b 2

3

4

are as above, then the vectors hab σ b , hab σ b , hab σ b are linearly independent. For, as we have just seen, 2

2

3

4

3

4

if a linear combination k (hab σ b ) + k (hab σ b ) + k (hab σ b ) of the three is 0, the three coefficients must all be 0.) So, at every point, hab [Vb ] is a three-dimensional subspace of the vector space of spacelike vectors. But the latter is itself three-dimensional. So every spacelike vector must be in hab [Vb ]. For (3), suppose hab σb = hab σ ′b . Then, by (1), (σ ′b − σb ) = k tb for some k. So hab σ ′a σ ′b =

hab (σa + k ta )(σb + k tb ) = hab σa σb .

1

So here is the indirect procedure. If µa is spacelike, we take its spatial length to be (hab σa σb ) 2 , where σb is a vector such that hab σb = µa . Clause (2) guarantees existence, and clause (3) guarantees that the choice of σb makes no difference. Proposition 4.1.1 has a number of simple consequences that will be used again and again in what follows. Here is one. Suppose we have a tensor γ...a... at a point such that (i) γ...a...hab = 0 and (ii) γ...a... ξ a = 0 for some timelike vector ξ a there. Then γ...a... = 0. (To see this, it suffices to consider any 2

3

4

three linearly independent spacelike vectors µ a , µ a , µ a at the point. (Existence is guaranteed by the signature of tab .) They, together with ξ a , form a basis for the tangent space there. Since we are given i

that γ...a... ξ a = 0, it suffices to show that γ...a...µa = 0 for i = 2, 3, 4. But we know from the proposition i

i

i

that, for each i = 2, 3, 4, there is a co-vector σ b such that µa = hab σ b . So our claim follows from (i).) This first consequence of Proposition 4.1.1 can be generalized. Suppose we have a tensor γ...ab... at a point such that, for some timelike vector ξ a there, (i) γ...ab... ham hbn = 0, and (ii) γ...ab...ξ a hbn = 0 = γ...ab... han ξ b , and (iii) γ...ab... ξ a ξ b = 0. Then γ...ab... = 0. Other tensors γ...a1 a2 ... an ... can be handled similarly. Proposition 4.1.2. Let (M, ta , hab , ∇) be a classical spacetime, and let ξ a be a smooth, future-directed, ˆ ab on unit timelike vector field on M . (So ta ξ a = 1.) Then there is a (unique) smooth, symmetric field h M satisfying the conditions ˆhab ξ b ˆ ab h h

bc

=

0,

=

δ ca

(4.1.10) c

− ta ξ .

(4.1.11)

ˆ ab Proof. If follows by the remark in the preceding paragraph that there can be at most one field h satisfying the stated conditions. (Given any two candidates, we need only substract one from the other ˆ ab by its and apply the remark to the resulting difference field.) We can define a symmetric field h specifying its action, at any point, on the unit timelike vector ξ a and on an arbitrary spacelike vector µa . ˆ ab that annihilates the former and makes the assignment So consider the field h ˆ ab µb = σa − ta (ξ c σc ) h

221


to the latter — where σa is any vector such that µa = hab σb . It is easy to check that the choice of σa 1

2

1

1

2

2

plays no role here. (For suppose that hab σ b = hab σ b . Then σ a − ta (ξ c σ c ) = σ a − ta (ξ c σ c ). The latter

follows since we get the same result on both sides if we contract with either ξ a or hab .) It now follows, as ˆ ab , both sides of (4.1.11) yield well, that condition (4.1.11) holds. For by the very way we have defined h

the same result when contracted with any vector σc . ˆ ab the spatial metric (or spatial projection field) relative to ξ a . Our notation is imperfect here We call h because we make no explicit reference to ξ a . But it will be clear from the context which unit timelike field is intended. Because hab is not invertible, we cannot raise and lower indices with it. But we can, at least, raise indices, and it is sometimes convenient to do so. So, for example, if Rabcd is the Riemann curvature tensor field associated with ∇, we can understand Rabcd to be the field hbn Rancd . Note that ˆ a = δ a − tb ξ a . h b b

(4.1.12)

ˆ a = ˆhmb hma .) It follows immediately from (4.1.12) that, given (This is simply equation (4.1.11), since h b any vector µa at a point, we can express it in the form ˆ a b µb + (tb µb ) ξ a . µa = h ˆ a b µb the Here the first term on the right side is spacelike, and the second is proportional to ξ a . We call h spatial projection (or spatial component) of µa relative to ξ a . ˆ ab µa µb ) 21 the spatial length of µa relative to ξ a . It is easy to check that this magnitude is We also call (h ˆ a b µb . (According just what we would otherwise describe as the spatial length of the spatial component h ˆ a b µb is given by (hmn σm σn ) 2 , where σm is any vector to our prescription, the spatial length of h ˆ mr µr . So the spatial length of h ˆ a b µb is given by ˆ a b µb = ham σm . But h ˆ a b µb = ham h satisfying h 1

1 ˆ mr µr ) (h ˆ ns µs ) 2 . hmn (h

ˆ ns = h ˆ rs . So the spatial length of ˆha b µb comes out as (h ˆ rs µr µs ) 12 , as claimed.) ˆ mr h But hmn h It is important that the compatibility conditions ∇a hbc = 0 and ∇a tb = 0 (or, equivalently, ∇a tbc =

0) do not determine a unique derivative operator. (There is no contradiction here with proposition 1.9.2 since neither tab nor hab is an (invertible) metric.) In fact, we have the following characterization result. Proposition 4.1.3. Let (M, ta , hab , ∇) be a classical spacetime. Let ∇′ = (∇, C abc ) be a second derivative

operator on M (i.e., the action of ∇′ relative to that of ∇ is given by C abc ). Then ∇′ is compatible with ta and hab iff C abc is of the form

C abc = 2 han t(b κc)n where κab is a smooth anti-symmetric field on M .

(4.1.13)

222


Proof. Since (M, ta , hab , ∇) is a classical spacetime, ∇ is compatible with both ta and hab . Hence, by

(1.7.1), we have

∇′a tb

=

∇a tb + C rab tr = C rab tr ,

(4.1.14)

∇′a hbc

=

∇a hbc − C bar hrc − C car hbr = −C bar hrc − C car hbr .

(4.1.15)

Assume, first, that C abc has the indicated form. Then ta C abc = 0 and C abc hcd = han tb κcn hcd by the orthogonality condition. It follows immediately that ∇′ is compatible with ta . It also follows that ∇′a hbc = −ta (hbn κrn hrc + hcn κrn hbr ) = −ta (κcb + κbc ). But κab is anti-symmetric. So ∇′ is compatible with hab as well. Conversely, assume ∇′ is compatible with ta and hab . Then, by (4.1.14) and (4.1.15), C rab tr

= 0,

(4.1.16)

C bar hrc + C car hbr

= 0.

(4.1.17)

Now consider the raised index tensor field C abc = C amn hmb hnc . It is spacelike, i.e., contraction on any index with ta yields 0. Moreover, it satisfies the two conditions C abc

= C acb ,

(4.1.18)

C abc

= −C cba .

(4.1.19)

(This first follows from the symmetry of C abc itself, and the second from (4.1.17).) By repeated use of these two, we have C abc = −C cba = −C cab = C bac = C bca = −C acb = −C abc . So the field vanishes everywhere: C abc = 0.

(4.1.20)

ˆ ab be the corresponding Now let ξ a be a smooth, future-directed, unit timelike field (so ta ξ a = 1), and let h spatial projection field. Then we have 0

=

ˆ nc ˆ mb h ˆ nc = C a hrm hsn h ˆ mb h C amn h rs

=

C ars (δ rb − tb ξ r )(δ sc − tc ξ s ).

Hence, C abc = tc C abs ξ s + tb C arc ξ r − tb tc C ars ξ r ξ s .

(4.1.21)

ˆ nq C p q ξ r + t[c h ˆ n]q C q ξ r ξ s . ˆ cp h κcn = −h r rs

(4.1.22)

Now consider


223

It is anti-symmetric by (4.1.17) and, we claim, it satisfies (4.1.13). To verify this, we compute the right side of (4.1.13 ). We have 2 han tb κcn

ˆ nq ) C q ξ r ξ s . ˆ cp C p q ξ r + tb tc (han h ˆ nq ) tb h = −2 (han h rs r

ˆ nq ) C q = C a . ˆ nq ) C p q = −(δ a − tq ξ a ) C q p = −C a p , and (han h Now, by (4.1.16) and (4.1.17), (han h rs rs r q r r So

2 han tb κcn

ˆ cp C a p ξ r + tb tc C a ξ r ξ s . = 2 tb h r rs

ˆ cp C a p = h ˆ cp C a hsp = (δ s − tc ξ s ) C a . So Furthermore, h r rs c rs 2 han tb κcn

=

2 tb C arc ξ r − tb tc C ars ξ r ξ s .

Hence, by (4.1.21), 2 han t(b κc)n = 2 t(b C ac)r ξ r − tb tc C ars ξ r ξ s = C abc . Now let Rabcd be the curvature tensor associated with ∇. Of course, it satisfies the algebraic conditions

listed in proposition 1.8.2:

Rab(cd)

=

0,

(4.1.23)

Ra[bcd]

=

0.

(4.1.24)

The compatibility conditions (∇a tb = 0 and ∇a hbc = 0) further imply that ta Rabcd

=

0,

(4.1.25)

(ab) R cd

=

0.

(4.1.26)

(We have 0 = 2∇[c ∇d] tb = ta Ra bcd and 0 = 2∇[c ∇d] hab = −Ramcd hmb − Rbmcd ham = −Rabcd − Rbacd .)

It follows immediately from the conditions listed so far that if we raise all three indices with hab , the resulting field Rabcd satisfies Rab(cd)

= 0,

(4.1.27)

Ra[bcd]

= 0,

(4.1.28)

R(ab)cd

= 0.

(4.1.29)

Rabcd = Rcdab .

(4.1.30)

These, in turn, jointly imply

(The argument is the same as in the case where ∇ is determined by a (non-degenerate) metric. Recall our proof of the fourth clause of proposition 1.9.4.)

Now consider the Ricci tensor field Rab = Rcabc and the (spatial) scalar curvature field R = hab Rab . We claim that the former is symmetric. To verify this, we consider an arbitrary smooth, future-directed,

224


ˆ ab to lower indices. First, it follows easily timelike field ξ a and use the corresponding projection field h from (4.1.11), (4.1.25), and (4.1.26) that Raacd R

bc

=

ˆ ab Rab = 0, h cd

(4.1.31)

=

ˆ ad R h

(4.1.32)

R =

abcd

,

ˆ ab Rab . h

(4.1.33)

ˆ ab Rab = h ˆ ab hbr Ra = (δ r − ta ξ r ) Ra = Ra . This, with (4.1.26), gives (For example, we have h a cd rcd rcd acd

(4.1.31).) Hence, by (4.1.23) and (4.1.24),

Rab − Rba = Rcabc − Rcbac = Rcabc + Rcbca = −Rccab .

(4.1.34)

Rab = Rba ,

(4.1.35)

So, by (4.1.31), we have

as claimed. Less straightforward is the following proposition. Proposition 4.1.4. Let (M, ta , hab , ∇a ) be a classical spacetime. Then the curvature field Rabcd associated with ∇ satisfies

Rabcd = (hbc Rad + had Rbc − hac Rbd − hbd Rac ) +

1 ac bd (h h − had hbc ) R. 2

(4.1.36)

Proof. The relation is familiar from the case where we are dealing with a derivative operator determined by an (invertible) metric and the background manifold has dimension 3. It follows from the symmetries (4.1.27) – (4.1.30) and (4.1.35), as well as the crucial fact that all the indices in Rabcd are spacelike, i.e., contraction on any of these indices with ta yields 0. We prove (4.1.36) at an arbitrary point p of M by introducing an appropriate basis there and considering the resulting component relations. 1

2

3

i

j

Let ta , σa , σa , σa be an orthonormal basis for hab at p in the sense discussed above. (So hab σ a σ b = 0 i

1

i

1

2

2

3

3

if i 6= j, and hab σ a σ b = 1 for i = 1, 2, 3.) Then hab = σ a σ b + σ a σ b + σ a σ b . Further, let ξ a be ˆ ab . Now consider the a future-directed unit timelike vector at p with corresponding projection tensor h 1

2

3

co-vectors αa , αa , αa at p defined by i ˆ ab hbc σi c = σi a − ta (σi c ξ c ). αa = h

It is easy to check that: i

(1) αa ξ a = 0 for i = 1, 2, 3. i

i

(2) hab σ a = hab αa for i = 1, 2, 3. 1

2

3

(3) ta , αa , αa , αa form a co-basis at p.

225


3 2 3 1 2 1 ˆ ab = α (4) h a αb + αa αb + αa αb .

Since all indices in Rabcd and Rab are spacelike, both tensors are determined by their action on the basis 1

2

3

vectors αa , αa , αa . Consider the components ij

R

Rab αa αb ,

=

Rabcd αa αb αc αd

ijkl

R

j

i

=

j

i

l

k

where i, j, k, l ∈ {1, 2, 3}. Because of the symmetries of Rabcd and Rab , each has only six independent (non-zero) components, namely

11

12

R

13

R

R

22

23

R

R

33

R

and 1212

R

ˆ rs Now, by (4.1.32), Rab = Rrabs h

1313

2323

1213

1223

1323

R R R R R . 3 ijki 3 X X jk i i R . This Rrabs αr αs . Hence, for all j, k ∈ {1, 2, 3}, R = = i=1

i=1

gives us

11

1212

1313

22

1212

2323

33

1313

2323

R = − R − R

12

1323

13

1223

23

1213

R = − R

R = − R − R

R =

R = − R − R

R

R = − R .

Also, by (4.1.33), ˆ ab = R = Rab h

3 X

i

i

11

22

33

Rab αa αb = R + R + R .

i=1

Using these relations, we can check that the two sides of (4.1.36) agree in their action on any quadruple i

j

k

l

1

2

1

2

1

1

2

2

1

2

αa αb αc αd . As an example, consider αa αb αc αd . We have hab αa αb = hab αa αb = 1 and hab αa αb = 0. 1212 22 11 1 So it suffices to confirm that R = (− R − R ) + R. But this follows from the entries in our table. 2 Next we consider the notion of “spatial flatness”. Of course, we say that our background classical spacetime is flat at a point if Rabcd = 0 there. In parallel, we say that it is spatially flat there if Rabcd = 0. To motivate this definition, we need to say something about “induced derivative operators” on spacelike hypersurfaces. (Recall that a hypersurface is spacelike — in a classical spacetime as well as in a relativistic spacetime — if all smooth curves with images in the hypersurface are spacelike.) Let S be a spacelike hypersurface, and let ξ a be an arbitrary smooth, unit, future-directed timelike ˆ ab be the associated projection field on S. Given any tensor field on S, we say vector field on S. Let h that it is spacelike relative to ξ a if contraction on any of its indices with ta or ξ a yields 0. We can think of fields spacelike relative to ξ a as living on the manifold S. (Recall the discussion in section 1.10.) Clearly, ˆ ab both qualify as spacelike relative to ξ a . So does h ˆ b = δ b − ta ξ b . Notice that h ˆ b preserves hab and h a

a

a

ˆ b µa = µb and h ˆ b σb = σa , for all µa and σa spacelike all vectors that are spacelike relative to ξ a , i.e., h a a ˆ b as a “delta (or index substitution) field” for fields on S that are relative to ξ a . We can thus think of h a

226


ˆ b – just as in the case of a spacelike relative to ξ a . And we shall, on occasion, write δˆba rather than h a (non-degenerate) metric gab we often write δ ba rather than g ba . ˆ ab as a (non-degenerate) metric that lives on What is most important here is that we can think of h S. It is non-degenerate in the relevant sense because it does not annihilate any non-zero vectors that ˆ ab hbc = δˆc . (This is are spacelike relative to ξ a or, equivalently, because it has an “inverse” hab , i.e., h a

ˆ ab , i.e., such that just 4.1.11.) So there is a unique derivative operator D on S that is compatible with h ˆ bc = 0. We can express the action of D in terms of ∇ (as explained in section 1.10). Given any Da h

field spacelike relative to ξ a , the action of D on it is given by first applying ∇ and then projecting all ˆ b . So, for example, covariant indices with h a ˆ r ˆhs ∇m αa . ˆm h Dn αabc = h c rs b n

(4.1.37)

The projection insures that the resultant field is spacelike relative to ξ a . There is no need to project the contravariant indices. Since ∇a tb = 0, they remain spacelike even after ∇ is applied. (One can

check directly that D satisfies all the defining conditions of a derivative operator on S, and furthermore ˆ bc = 0 and Da hbc = 0.) We refer to D as the derivative operator induced on S relative to ξ a . Da h The following proposition serves to motivate our definition of spatial flatness.

Proposition 4.1.5 (Spatial Flatness Proposition). Let (M, ta , hab , ∇) be a classical spacetime. The

following conditions are equivalent at every point in M . (1) Space is flat, i.e., Rabcd = 0. (2) Rab = 0. (3) Rab = t(a ϕb) for some ϕa .

Furthermore, given any spacelike hypersurface S in M , these conditions hold throughout S iff parallel transport of spacelike vectors within S is, at least locally, path independent. Proof. Let p be a point in M , and let ξ a be an arbitrary, future-directed, unit timelike vector at p with ˆ ab . The equivalence of (1) and (2) follows from (4.1.32), (4.1.33), corresponding spatial projection tensor h and (4.1.36). The implication (3) ⇒ (2) is immediate. For the converse, consider the vector ϕa = 2 Rab ξ b − ta (Rmn ξ m ξ n ). We have ϕa ξ a = Rab ξ a ξ b and ϕa har = 2Rab har ξ b . Therefore, at any point where Rab = 0, it must be the case that Rab = t(a ϕb) , since both sides agree in their action on ξ a ξ b , har ξ b , and har hbs . (Recall our remarks following proposition 4.1.1.) Now let S be a spacelike hypersurface, and let ξ a be a smooth, unit, future-directed timelike vector ˆ ab be the associated projection field on S, and let D be the derivative operator field on S. Further, let h induced on S relative to ξ a (as explained in the preceding paragraphs). Finally, suppose that µa and ν a are spacelike fields on S. Then they automatically qualify as spacelike relative to ξ a , and by (4.1.37) we

227


ˆ r ∇r ν a = µr ∇r ν a . It follows that D and ∇ induce the same conditions for parallel have µn Dn ν a = µn h n

transport of spacelike vectors on S. We know that parallel transport of such vectors on S is, at least

locally, path independent iff the Riemann curvature tensor field Rabcd on S associated with D vanishes.

So, for the second half of the proposition, it suffices for us to show that, at all points on S, Rabcd = 0 ⇐⇒ Rabcd = 0.

(4.1.38)

This just involves a bit of computation. The right-side condition here is equivalent to the requirement that, for all spacelike fields µa on S, ˆ s Ra µb . ˆ s ∇[r ∇s] µa = h ˆr h 0 = Rabcd µb = −2 D[c Dd] µa = −2 ˆhr c h d brs c d Hence, it is equivalent to the condition ˆ s Rab . ˆr h ˆ s Ra hpb = h ˆr h 0 = h rs d d prs c c Contracting this equation with hcm hdn yields Rabmn = 0. Conversely, contracting Rabmn = 0 with ˆ s Rab = 0. ˆr h ˆ cm ˆhdn yields h h rs c d The interest of proposition 4.1.5 will become apparent in the next section when we consider the geometrized formulation of Newtonian gravitation theory. In that formulation, Poisson’s equation assumes the form Rab = 4 πρ ta tb (where ρ is the mass density function). We see from the proposition that Poisson’s equation (in its geometrized formulation) implies the flatness of space! This is striking. It is absolutely fundamental to the idea of geometrized Newtonian theory that spacetime is curved (and gravitation is just a manifestation of that curvature). Yet the basic field equation of the theory itself rules out the possibility that space is curved. Intermediate between the curvature conditions Ra bcd = 0 and Rabcd = 0 is the condition Rab cd = 0. We shall show later (proposition 4.3.1) that it holds throughout M iff parallel transport of spacelike vectors along arbitrary curves is, at least locally, path independent. (Here we still restrict attention to spacelike vectors (rather than arbitrary vectors), but consider their transport along arbitrary curves in M (not just curves confined to a particular spacelike hypersurface).) Before continuing with the main line of presentation in this section, we stop briefly to record a fact that will be needed in later sections. We place it here because it concerns the induced derivative operator D that was considered in the preceding proof. Proposition 4.1.6. Let (M, ta , hab , ∇) be a classical spacetime, and let φa be smooth spacelike field on

M such that ∇[a φb] = 0. Then, at least locally, there exists a smooth field φ such that φa = ∇a φ.

Proof. This is not quite an instance of proposition 1.8.3, but it is close. Let p be any point in M , and let O be any open set containing p that is sufficiently small and well behaved that it has this property: O can be covered by a family F of spacelike hypersurfaces, each of which is connected and simply connected.


228

Let γ : I → M be any timelike curve whose image contains p and intersects every one of the hypersurfaces ˆ ab be the associated in F . Finally, let ξ a be a smooth, future-directed, unit timelike field on O, and let h spatial projection field.

Now consider any hypersurface S in F , and the projected field φâ = ˆhab φb on S. If D is the induced ˆ bn ∇[m φn] = 0. So, by ˆ am h derivative operator on S defined by (4.1.37), then on S we have D[a φˆb] = h proposition 1.8.3, there is a smooth field φS on S such that φâ = Da φS . It is determined only up to a constant, but we can pin it down uniquely by requiring, in addition, that it have value 0 at the point where S intersects γ[I]. Now let φ be the “aggregated” scalar field on O that agrees with φS on each S in F . We claim

without further argument that it is smooth. It satisfies the required condition since, given any spacelike ˆ r n ∇r φ = ∇a φ on S. hypersurface S in F , we have φa = han φˆn = han Dn φS = han h Now we briefly consider the representation of fluid flow. Our formalism here is closely related to that developed in section 2.8. Let ξ a be a smooth, unit, future-directed timelike vector field on our ˆ ab be the projection background classical spacetime. We think of ξ a as the four-velocity of a fluid. Let h field associated with ξ a . The rotation field ωab and expansion field θab associated with ξ a are defined by ωab

=

ˆ m [a h ˆ b]n ∇m ξ n , h

(4.1.39)

θab

=

ˆ m (a h ˆ b)n ∇m ξ n . h

(4.1.40)

(We can motivate the terminology here much as we did in section 2.8.) It follows that ˆ bn ξ m ∇m ξ n ˆ bn ∇a ξ n = ωab + θab + ta h h

(4.1.41)

∇a ξ b = ωab + θab + ta ξ m ∇m ξ b

(4.1.42)

∇a ξ b = ω ab + θab .

(4.1.43)

and, hence, that

and

As in the relativistic case, we can decompose the expansion field to arrive at the scalar expansion field θ and the shear field σab : θ

=

σab

=

θa a = ∇a ξ a , 1 ˆ θab − θ h ab . 3

(4.1.44) (4.1.45)

(That θa a = ∇a ξ a follows from (4.1.42) and the anti-symmetry of ωab .) Clearly, σab is “trace-free” since ˆ a = θ − 1 θ(δ a − ta ξ a ) = θ − 1 θ(4 − 1) = 0. We note for future reference the following σ a = θ a − 1 θh a

a

3

a

3

a

3

equivalences:

ωab = 0

⇐⇒

∇[a ξ b] = 0,

(4.1.46)

θab = 0

⇐⇒

∇(a ξ b) = 0.

(4.1.47)

229


(In each case, we get the implication from left to right by raising indices with hmn , and the one from ˆ mn .) The conditions in the first line capture the claim that ξ a is right to left by lowering indices with h non-rotating (or twist-free). Finally, we say just a bit about the four-momentum of point particles and the four-momentum density of matter fields. It is instructive to consider the situations in Newtonian and relativistic mechanics side by side. (For a more complete and thorough comparison, see Dixon [12].) Suppose, first, that we have a point particle with mass m and four-velocity field ξ a . Then, just as in relativity theory, we associate with it a four-momentum field P a = m ξ a along its worldline. (In the present context we have only particles with positive mass (m > 0) to consider.) Suppose particle O has four-velocity ξ a at a point, and another particle O′ has four-momentum P a = m ξ ′a there. Just as in the relativistic case, we can decompose P a relative to ξ a . Newtonian Mechanics P a = (tb P b ) ξ a | {z } mass

+

ˆ ab P b h } | {z

relative 3−momentum

Relativistic Mechanics Pa =

(ξb P b ) ξ a + | {z }

relative energy

ha b P b | {z }

relative 3−momentum

But the decomposition works somewhat differently in the two cases. In Newtonian mechanics, we have a component proportional to ξ a with magnitude tb P b = tb (m ξ ′b ) = m, and a spacelike component ˆ a b P b = (δ a b − tb ξ a ) (m ξ ′b ) = m (ξ ′b − ξ a ), h which gives the three-momentum of the particle relative to ξ a . (The vector (ξ ′b − ξ a ) by itself gives the relative velocity of O′ with respect to O.) Thus, in Newtonian mechanics, the four-momentum P a

of a point particle codes its mass and its three-momentum, as determined relative to other background observers. So it is appropriately called the “mass-momentum vector”. In relativistic mechanics, in contrast, as we have seen, the component of P a proportional to ξ a has magnitude (ξb P b ), which gives the energy of the particle as determined relative to ξ a . And we call P a the “energy-momentum vector”. In relativistic mechanics, the mass of the particle is given by the length of its four-momentum 1

(gab P a P b ) 2 . The corresponding statement in Newtonian mechanics is that the mass of the particle 1

is given by the temporal length of its four-momentum (tab P a P b ) 2 . Now we switch from point particles to continuous matter fields. Just as in relativity theory, we associate with each matter field F a smooth, symmetric field T ab . But the interpretation of T ab is

different in Newtonian mechanics (parallel to the way that the interpretation of P a is different), and here we call it the mass-momentum field associated with F . In both cases, T ab codes the four-momentum

density of F as determined, at any point, relative to future-directed, unit timelike vectors ξ a there. But

in the Newtonian case, the four-momentum density is the same for all ξ a . It is given by T ab tb . (In the relativistic case, it is not invariant and is given, instead, by T ab ξb . Recall section 2.5.)

230


Newtonian Mechanics

Relativistic Mechanics

T ab tb is the four-momentum density of F

T ab ξb is the four-momentum density of F as determined relative to ξ a

The conservation equation carries over intact from relativistic mechanics: ∇a T ab = 0.

(4.1.48)

We can decompose the Newtonian four-momentum density T ab tb just as we decomposed P a to determine an invariant mass-density and a relative three-momentum density. The former is given by ρ = T ab ta tb . We can take it to be a (Newtonian) “mass condition” that T ab ta tb > 0 whenever T ab 6= 0. When the condition is satisfied, we can further define the fields ηa

=

pab

=

1 ab T tb , ρ T ab − ρ η a η b ,

and arrive at a canonical representation of T ab : T ab = ρ η a η b + pab .

(4.1.49)

Here η a is a smooth, future-directed, unit timelike field, and pab is a smooth, symmetric field that is spacelike in both indices (ta pab = 0). In the case of a fluid, for example, we can interpret η a as the four-velocity of the fluid. In terms of this representation, the conservations equation comes out as 0 = ∇a T ab = ρ η a ∇a η b + η b [η a ∇a ρ + ρ ∇a η a ] + ∇a pab .

(4.1.50)

Contracting with tb yields the following equivalence:   ρ η a ∇ η b + ∇ pab = 0 a a ∇a T ab = 0 ⇐⇒  η a ∇a ρ + ρ (∇a η a ) = 0.

The second equation on the right expresses the conservation of mass. (The analysis we gave in the

context of relativity theory carries over intact.) The first is an equation of motion. In the case of a perfect fluid, for example, pab = p hab , where p is the (isotropic) pressure of the fluid. In this case, the first equation comes out as Euler’s equation: ρ η a ∇a η b = −∇b p.

(4.1.51)

For more on the development of Newtonian mechanics within our geometric framework, see, for example, Ellis [17] and K¨ unzle [35].


4.2

231

Geometrized Newtonian Theory — First Version

Now we turn to Newtonian gravitation theory proper. In the standard (non-geometrized) version, one assumes that the background derivative operator ∇ is flat and posits a gravitational potential φ.

The gravitational force on a point particle with mass m is given by − m hab ∇b φ. (Notice that this is a

spacelike vector by the orthogonality condition.) Using our convention for raising indices, we can also express the vector as − m ∇a φ. It follows that if the particle is subject to no forces except gravity, and if it has four-velocity ξ a , it satisfies the equation of motion

−∇a φ = ξ n ∇n ξ a .

(4.2.1)

(Here we have just used − m ∇a φ for the left side of (4.1.9).) It is also assumed that φ satisfies Poisson’s

equation

∇a ∇a φ = 4 π ρ,

(4.2.2)

where ρ is the Newtonian mass-density function. (The expression on the left side is an abbreviation for hab ∇a ∇b φ.) In the geometrized formulation of the theory, gravitation is no longer conceived of as a fundamental “force” in the world but rather as a manifestation of spacetime curvature, just as in relativity theory. Rather than thinking of point particles as being deflected from their natural straight trajectories in flat spacetime, one thinks of them as traversing geodesics in curved spacetime. So we have a geometry problem. Starting with a classical spacetime (M, ta , hab , ∇), with ∇ flat and with field φ on M , can g

we find a new derivative operator ∇ on M , also compatible with ta and hab , such that a timelike curve

satisfies the equation of motion (4.2.1) with respect to the original derivative operator ∇ iff it is a geodesic g

with respect to ∇? The following proposition (essentially due to Trautman [59]) asserts that there is g

exactly one such ∇. It also records several conditions satisfied by the Riemann curvature tensor field g

g

R a bcd associated with ∇. We shall consider the geometric significance of these conditions in section 4.3.

Proposition 4.2.1 (Geometrization Lemma). Let (M, ta , hab , ∇) be a classical spacetime with ∇ flat

(Ra bcd = 0). Further, let φ and ρ be smooth real valued functions on M satisfying Poisson’s equation g

∇a ∇a φ = 4 π ρ. Finally, let ∇= (∇, C abc ), with C abc = −tb tc ∇a φ. Then all the following hold. g

(G1) (M, ta , hab , ∇) is a classical spacetime. g

(G2) ∇ is the unique derivative operator on M such that, for all timelike curves on M with four-velocity field ξ a ,

g

g

(G3) The curvature field R

a

ξ n ∇n ξ a = 0 ⇐⇒ ξ n ∇n ξ a = −∇a φ.

(4.2.3)

g

bcd

associated with ∇ satisfies g

= 4 π ρ tb tc ,

Rbc g

R g

R

g

(4.2.4)

a c b d

= R cdab,

(4.2.5)

ab

= 0.

(4.2.6)

cd

232


g

Proof. For (G1), we need to show that ∇ is compatible with tb and hab . But this follows from proposition 4.1.3, for we can express C abc in the form C abc = 2 han t(b κc)n if we take κcn = −t[c ∇n] φ. g

For (G2), let ∇= (∇, C abc ) where (temporarily) C abc is an arbitrary smooth symmetric field on M . Let

p be an arbitrary point in M , and let ξ a be the four-velocity field of an arbitrary timelike curve through p. Then, by (1.7.1),

g

ξ n ∇n ξ a = ξ n ∇n ξ a − C arn ξ r ξ n .

g

It follows that ∇ will satisfy (G2) iff C arn ξ r ξ n = −∇a φ or, equivalently, [C arn + (∇a φ) tr tn ] ξ r ξ n = 0

(4.2.7)

for all future-directed unit timelike vectors ξ a at all points p. But the space of future-directed unit 1

4

timelike vectors at any p spans the tangent space Mp there. (Why? Let ξ a , ..., ξ a be an orthonormal 1

i

basis for tab = ta tb in the sense already discussed here. (So ta ξ a = 1, and ta ξ a = 0 for i = 2, 3, 4.) Then 1

1

2

1

3

1

4

ξa , (ξa + ξa ), (ξa + ξa ), and (ξa + ξa ) are all future-directed unit timelike vectors, and the set is linearly g

independent.) And the field in brackets in (4.2.7) is symmetric in its covariant indices. So, ∇ will satisfy

(G2) iff C arn = −(∇a φ) tr tn everywhere.

Finally, for (G3) we use (1.8.2). We have g

R

a

bcd

=

Ra bcd + 2 ∇[c C a d]b + 2 C n b[c C a d]n

=

Ra bcd − 2 tb t[d ∇c] ∇a φ = − 2 tb t[d ∇c] ∇a φ.

(4.2.8)

(Here C n b[c C a d]n = 0 by the orthogonality condition, and ∇[c C a d]b = −tb t[d ∇c] ∇a φ by the compatibility

condition. For the final equality, we use our assumption that Rabcd = 0.) (4.2.6) now follows from the

orthogonality condition. (4.2.5) follows from that and the fact that ∇[c ∇a] φ = 0 for any smooth function

φ. Finally, contraction on a and d yields

g

a R bc = tb tc (∇a ∇ φ).

(4.2.9)

So (4.2.4) follows from our assumption that ∇a ∇a φ = 4 π ρ. (4.2.4) is the geometrized version of Poisson’s equation. In the special case where ρ = 0 everywhere, g

of course, it reduces to Rbc = 0, which we recognize as Einstein’s equation in the corresponding special case in which Tbc = 0. Even in the general case, (4.2.4) can be reformulated so as to have almost exactly the same structure as Einstein’s equation. Recall our discussion of the mass-momentum field T ab toward the end of section 4.1. We saw there that it encodes ρ via ρ = T mn tmn . (We shall temporarily revert to writing tab , rather than ta tb , to emphasize the field’s relation to a two index Lorentzian metric gab , but nothing turns on our doing so.) So we can certainly formulate Poisson’s


233

equation directly in terms of T ab . Now consider the fields Tˆbc

=

T mn tmb tnc = ρ tbc ,

Tˆ

=

T mn tmn = ρ.

(Caution is required here. It must be remembered that we cannot recover T bc from Tˆbc by “raising indices” with hab , since Tˆmn hmb hnc = 0.) Using these fields, we can express Poisson’s equation in the form g

Rbc = 8 π (Tˆbc −

1 tbc Tˆ), 2

(4.2.10)

which is very close indeed to Einstein’s equation (2.7.2). Moreover, if we start with a version of Poisson’s equation that incorporates a “cosmological constant” ∇a ∇a φ + Λ = 4 π ρ,

(4.2.11)

then substitution for ∇a ∇a φ in (4.2.9) yields g

Rbc = 4 π ρ tb tc − Λ tb tc

(4.2.12)

(but everything else in the proof remains the same). And this equation, in turn, can be expressed as g 1 Rbc = 8 π (Tˆbc − tbc Tˆ) − Λ tbc , 2

(4.2.13)

which matches (2.7.4). So far, we have seen how to pass from a standard to a geometrized formulation of Newtonian theory. It is also possible to work in the opposite direction. In Trautman’s [59] version of geometrized Newtonian gravitation theory — one of two we shall consider2 — one starts with a curved derivative operator ∇

satisfying (4.2.4), (4.2.5), (4.2.6), and with the principle that point particles subject to no forces (except “gravity”) traverse geodesics with respect to ∇. (4.2.5) and (4.2.6) function as integrability conditions

that ensure the possibility of working backwards to recover the standard formulation in terms of a gravf

itational potential φ and flat derivative operator ∇. We shall prove this recovery, or de-geometrization,

theorem in this section (proposition 4.2.5), and we shall see that, in the absence of special boundary f

conditions, the pair (∇, φ) that one recovers is not unique. Later, in section 4.5, we shall consider a second, more general version of geometrized Newtonian gravitation theory, developed by K¨ unzle [34, 35] and Ehlers [15], in which one of the two supplemental curvature conditions is dropped.

Trautman Version

2 See

    Rbc   

Ra b c d

Rab cd

 = 4 π ρ tb tc  K¨ unzle-Ehlers Version = Rc d a b 

=

0

Bain [4] for a systematic discussion of these and yet other versions.

234


At issue here is whether “Newtonian gravitation theory” is to qualify as a limiting version of relativity theory. The geometrized version of Poisson’s equation does, in a natural sense, qualify as a limiting form of Einstein’s equation. And the first of Trautman’s two supplemental curvature conditions (Ra b c d = Rc d a b ) holds automatically in relativistic spacetimes. (Recall the fourth clause of proposition 1.9.4.) So it naturally carries over to any limiting version of relativity theory. But the second supplemental curvature condition does not hold in relativistic spacetimes (unless they happen to be flat), and it is therefore not an automatic candidate for inclusion in a limiting version of relativity theory. It is crucially important that the conditions Rab cd = 0 and Ra bcd = 0 are not equivalent for classical spacetime structures, though they are for relativistic ones. Starting only from the weaker assumptions of K¨ unzle and Ehlers, one can still prove a recovery theorem of sorts. But the (de-geometrized) gravitation theory one recovers is not Newtonian theory proper, but rather a generalized version of it. In this version, the gravitational force acting on a particle of unit mass is given by a vector field, but it need not be of the form ∇a φ. Moreover, the de-geometrized field equations

to which one is led involve a “rotation field” ωab . We shall eventually prove this recovery theorem for the

K¨ unzle-Ehlers version of Newtonian theory (proposition 4.5.2), and also consider special circumstances under which the difference between the two versions of geometrized Newtonian theory collapses. It turns out that the second curvature condition (Rab cd = 0) is satisfied automatically, for example, in classical spacetimes that are, in a certain weak sense, asymptotically flat (see section 4.5), and also in Newtonian cosmological models that satisfy a natural homogeneity and isotropy condition (see section 4.4). Before turning to the Trautman Recovery Theorem, we isolate a few needed facts. Let ξ a be a smooth, future-directed, unit timelike field in a classical spacetime (M, ta , hab , ∇). We say that it is rigid

(or non-expanding) if £ξ hab = 0 or, equivalently, ∇(a ξ b) = 0. (These conditions obtain, we know, iff

the expansion field θab associated with ξ a vanishes. Recall (4.1.47).) Certain things we have established about Killing fields (which we have defined only in connection with non-degenerate metrics) carry over to rigid fields in classical spacetimes. So, for example, we have the following. Proposition 4.2.2. Let (M, ta , hab , ∇) be a classical spacetime, and let ξ a be a smooth, future-directed,

unit timelike field that is rigid. Then

∇n ∇a ξ b = Rba r n ξ r .

(4.2.14)

Proof. The proof is a just a variant of that used for proposition 1.9.8. Cycling indices, we have ∇n ∇a ξ b − ∇a ∇n ξ b

= −Rb r na ξ r ,

∇b ∇n ξ a − ∇n ∇b ξ a

= −Ra r bn ξ r ,

∇a ∇b ξ n − ∇b ∇a ξ n

= −Rn r ab ξ r .

Subtracting the third line from the sum of the first two (and using the fact that ∇(a ξ b) = 0) yields 2 ∇n ∇a ξ b = (−Rb r na − Ra r bn + Rn r ab ) ξ r .


235

Finally, we reformulate the expression in parentheses on the right side using the symmetries Ra [bcd] = 0, Rab (cd) = 0, and R(ab) cd = 0: −Rb r na − Ra r bn + Rn r ab

=

(Rba r n + Rbna r ) − Ra r bn + Rn r ab

=

Rba r n + (Rnb r a + Rn r ab ) − Ra r bn

=

Rba r n − Rnab r − Ra r bn

=

Rba r n − (Ran r b + Ra r bn )

=

Rba r n + Rabn r = 2 Rba r n .

So we have (4.2.14). Our proof of the Trautman Recovery Theorem turns on the availability of a unit timelike field η a that is rigid and twist-free (∇a η b = 0). The latter provides a backbone, of sorts, for our construction. The following proposition shows that the condition Rab cd = 0 insures the existence of such fields (at least locally). Proposition 4.2.3. Let (M, ta , hab , ∇) be a classical spacetime that is spatially flat (Rabcd = 0). Let

γ : I → M be a smooth, future-directed timelike curve with unit tangent field ηb a , and let p be any point

in γ[I]. Then there is an open set O containing p, a smooth spacelike field χa on O, and a smooth, future-directed, unit timelike field η a on O such that χa = 0 on γ[I], η a = ηb a on γ[I], and ∇a χb = δa b − ta η b .

(4.2.15)

Furthermore, (i) if Rab cd = 0, then ∇a η b = 0; and (ii) if Ra bcd = 0 and if γ is a geodesic, then ∇a η b = 0. Proof. First, we claim there exists a smooth spacelike field χa on some open set O containing p such that ∇a χb = hab

(4.2.16)

and χa = 0 on γ[I]. Indeed, as restricted to any one spacelike hypersurface S, χa is just the familiar “position vector field” (relative to the point where γ[I] intersects S). (Recall proposition 1.7.12. All we need here is that the (three-dimensional, invertible) metric gab induced on S by hab is flat and so, at least locally, the pair (S, gab ) is isometric to three-dimensional Euclidean space.) Now let ξ a be any smooth, future-directed, unit timelike field on O. Consider the field η b = (−ξ a ∇a χb + ξ b ). We claim that it satisfies all the required conditions. First, it satisfies (4.2.15). This follows since the two fields

(−∇a χb + δa b ) and ta η b yield the same result when contracted with either hna or ξ a . Next, it is clearly a future-directed, unit timelike field, i.e., tb η b = tb (−ξ a ∇a χb + ξ b ) = 1. Third, it agrees with ηb a on γ[I].

For since χa vanishes on γ[I], it is certainly constant along the curve, i.e., ηb a ∇a χb = 0. So, by (4.2.15),

we have 0 = ηb a ∇a χb = ηb a (δa b − ta η b ) = ηb b − η b on γ[I].

Now we turn to the curvature conditions. By (4.2.15) again, η a ∇n ∇a χb = η a ∇n (δa b − ta η b ) = −∇n η b .

236


Hence, ∇n η b = −η a (∇a ∇n χb − Rb m n a χm ) = η a Rb m n a χm , since, by (4.2.16), ∇a ∇n χb = ∇a hnb = 0. Since χa is spacelike, we can express it in the form χa = hab χ ˆb .

Thus we have

∇n η b = Rbmn a χ ˆm η a .

(4.2.17)

So, if Rbm na = 0, it clearly follows that ∇n η b = 0. This gives us (i). Now assume that γ is a geodesic and Ra bcd = 0. Then ∇a η b = 0 on γ[I]. (Why? η a ∇a η b = 0 on γ[I]

since γ is a geodesic, and hna ∇a η b = 0 everywhere by (i).) We may assume (by moving to a smaller open

set O containing p if necessary) that every maximally extended spacelike hypersurface in O intersects

γ[I]. So it will suffice for (ii) to show that ∇a η b is constant on spacelike hypersurfaces, i.e., ∇c ∇a η b = 0. But this follows immediately from Ra bcd = 0 and ∇a η b = 0, since ∇c ∇a η b = ∇a ∇c η b − Rb m c a η m .

Proposition 4.2.3 yields a useful characterization of the relative strength of two curvature conditions. (Here and throughout it should be understood that when we formulate a curvature equation without qualification, as on the left sides of (1) and (2) that follow in proposition 4.2.4, we have in mind the condition that the equation hold at all points.) Proposition 4.2.4. Let (M, ta , hab , ∇) be a classical spacetime that is spatially flat (Rabcd = 0). Then the following both hold.

(1) Rab cd = 0 iff there exists, at least locally, a smooth unit timelike field η a that is rigid and twist-free (∇a η b = 0). (2) Ra bcd = 0 iff there exists, at least locally, a smooth unit timelike field η a that is rigid, twist-free, and acceleration-free (∇a η b = 0). Proof. The “only if” clauses follow from the preceding proposition. The other drections are easy. (1) Assume that for any point p in M , there exists a smooth unit timelike field η a defined on an open set containing p such that ∇a η b = 0. We show that Rab cd vanishes at p. We have Rab cd η c η d = 0 at

p since Rab cd is anti-symmetric in the indices c and d. We also have Rab cd hcr hds = 0 at p (by our

assumption of spatial flatness). So to prove that Rab cd vanishes at p, it suffices to show that contraction there with η c hds (or hcr η d ) yields 0. But this follows since ∇b η a = 0 and hence, by proposition 4.2.2,

Rab cd η c hds = ∇s ∇b η a = 0.

(2) Next, assume that for any point p in M , there exists a smooth unit timelike field η a defined on an open set containing p such that ∇a η b = 0. We show that Ra bcd vanishes at p. We know from part (1) of the proposition (and the fact that ∇a η b = 0 implies ∇a η b = 0) that Rab cd = 0 at p. The latter

condition implies that Ra bcd = tb Ra ncd η n . (Contracting both sides with either hbr or η b yields the same result.) But since ∇a η d = 0, we also have Ra ncd η n = −2 ∇[c ∇d] η a = 0 at p. So we are done.

237


Now we turn to our first recovery theorem. Our formulation is purely local in character since we have opted not to impose special global topological constraints on the underlying manifold M . Our proof is a bit different from that in Trautman [59]. Proposition 4.2.5. (Trautman Recovery Theorem) Let (M, ta , hab , ∇) be a classical spacetime that sat-

isfies

Rbc

= 4 π ρ tb tc ,

(4.2.18)

Ra b c d = Rc d a b ,

(4.2.19)

Rab cd = 0

(4.2.20)

for some smooth scalar field ρ on M . Then given any point p in M , there is an open set O containing p, f

a smooth scalar field φ on O, and a derivative operator ∇ on O such that all the following hold on O. f

(R1) ∇ is compatible with ta and hab . f

(R2) ∇ is flat.

(R3) For all timelike curves with four-velocity field ξ a , f

f

ξ n ∇n ξ a = 0 ⇐⇒ ξ n ∇n ξ a = − ∇ a φ.

f

f

(4.2.21)

f

(R4) ∇ satisfies Poisson’s equation ∇ a ∇ a φ = 4 π ρ. f

f

The pair (∇, φ) is not unique. A second pair (∇ ′ , φ′ ) (defined on the same open set O) will satisfy the

stated conditions iff

(U1) ∇a ∇b (φ′ − φ) = 0, and f

f

(U2) ∇′ = (∇, C ′a bc ), where C ′a bc = tb tc ∇a (φ′ − φ).

Proof. Let p be a point in M . As we have just seen, it follows from Rab cd = 0 that we can find an open set O containing p as well as a smooth, future-directed, unit timelike vector field η a on O that is rigid and twist-free (∇a η b = 0). Let φa be the acceleration field of η a , i.e., φa = η n ∇n η a . Then we have ∇a η b = ta φb .

(4.2.22)

(This follows since contraction of the two sides with both η a and han yields the same result.) Further, let f

f

∇ be the derivative operator on O defined by ∇ = (∇, C a bc ), where C a bc = tb tc φa . Clearly, ta C a bc = 0, C a bc hbn = 0, and C a bc hcn = 0. It follows that f

∇a tb

f

bc ∇a h

=

∇a tb + tn C n ab = ∇a tb ,

=

∇a hbc − hnc C b na − hbn C c na = ∇a hbc . f

So, since ∇ is compatible with ta and hbc , ∇ is compatible with them as well. So we have (R1). Notice

next that C b an η n = ta φb and so, by (4.2.22), f

b b b n b b ∇a η = ∇a η − C an η = ta φ − ta φ = 0.

(4.2.23)

238


f

Thus, η a is constant with respect to the new derivative operator ∇. f

Now we consider the curvature field associated with ∇. We have C n bc C a dn = 0 since φn tn = 0. So,

by (1.8.2),

f

R

a

= Ra bcd + 2 ∇[c C a d]b + 2 C n b[c C a d]n

bcd

= Ra bcd + 2 tb t[d ∇c] φa .

(4.2.24)

f

f

It follows immediately that Rabcd = Rabcd = 0 (since Rab cd = 0). So ∇ is spatially flat. But now recall the second clause of proposition 4.2.4. We have just verified that there is smooth unit timelike field η a f

f

f

on O that is constant with respect to ∇. So (since ∇ is spatially flat), the proposition tells us that ∇ f

must be flat outright, i.e., Ra bcd = 0. So we have (R2). And (4.2.24) reduces to Ra bcd

=

−2 tb t[d ∇c] φa .

(4.2.25)

Now we extract further information from (4.2.25). Raising and contracting indices yields Ra b c d Rbc

= −tb td ∇c φa ,

(4.2.26)

= tb tc ∇a φa .

(4.2.27)

Since we are assuming Ra b c d = Rc d a b , it follows from the first of these assertions that ∇[c φa] = 0.

This implies that (after possibly further restricting O to some smaller open set containing p) there is

a smooth scalar field φ on O such that φa = ∇a φ. (Here we invoke proposition 4.1.6.) And since we

are assuming Rbc = 4 π ρ tb tc , it follows from the second assertion that ∇a ∇a φ = ∇a φa = 4 πρ. But

C a an = ta tn φa = 0 and, therefore, f

f

f

f

a a a n a ∇ a ∇ φ = ∇a ∇ φ − C an ∇ φ = ∇a ∇ φ. f

(4.2.28)

f

So ∇ a ∇ a φ = 4 πρ. That is, we have (R4). For (R3), note that for all timelike curves in O with four-velocity field ξ a , f

ξ n ∇n ξ a

ξ n (∇n ξ a − C a nm ξ m ) = ξ n ∇n ξ a − (tn tm ∇a φ) ξ n ξ m = ξ n ∇n ξ a − ∇a φ.

=

f

So ξ n ∇n ξ a = 0 iff ξ n ∇n ξ a = −∇a φ.

f

f

Finally, we consider the non-uniqueness of the pair (∇, φ). Let (∇ ′ , φ′ ) be a second pair on O. f

f

Consider fields C ′a bc and ψ on O defined by ∇′ = (∇, C ′a bc ) and ψ = φ′ − φ. We first show that if the

new pair satisfies the stated conditions of the proposition, then it must be the case that ∇a ∇b ψ = 0 and

C ′a bc = tb tc ∇a ψ.

f

f

f

Assume (∇′ , φ′ ) satisfies (R1) - (R4). Then — since (∇, φ) and (∇′ , φ′ ) both satisfy (R3) — we have f

f

f

f

ξ n ∇n ξ a + ∇ a φ = 0 ⇐⇒ ξ n ∇n ξ a = 0 ⇐⇒ ξ n ∇′ n ξ a + ∇′a φ′ = 0

239


f

f

f

f

f

for all timelike curves with four-velocity field ξ a . But ∇′a φ′ = ∇a φ′ = ∇a φ + ∇a ψ. And ξ n ∇′ n ξ a = f

ξ n (∇n ξ a − C ′a nm ξ m ). So it must be the case that, for all future-directed, unit timelike vectors ξ a at all

points in O,

f

C ′a nm ξ m ξ n = ∇ a ψ. f

And from this it follows that C ′a mn = tm tn ∇ a ψ = tm tn ∇a ψ, as required. (Recall the argument for a f

corresponding assertion in our proof of the Geometrization Lemma.) Now the curvature fields of (∇, φ) f

and (∇′ , φ′ ) are related by

f

R

′a

f

bcd

f

f

= R a bcd + 2 tb t[d ∇c] ∇ a ψ. f

f

(The argument here is exactly the same as given above for (4.2.24).) Since ∇ and ∇ ′ are both flat, it f

f

f

f

f

f

f

follows that t[d ∇c] ∇ a ψ = 0 or, equivalently, ∇ c ∇ a ψ = 0. But ∇c ∇a ψ = ∇ c ∇ a ψ. (Indeed, ∇ c f

and ∇c agree in their action on all vector fields λa , since ∇ c λa = ∇c λa − C ac n λn and C ac n = 0.) So ∇c ∇a ψ = 0, and we are done with the first direction. f

Conversely, assume that C ′a bc = tb tc ∇a ψ and ∇a ∇b ψ = 0. The first assumption alone implies that

∇′ is compatible with ta and hab . And by reversing the steps in the preceding paragraphs, we can show f

that (∇′ , φ′ ) satisfies (R2) and (R3). That leaves only (R4). For this, note first that since C ′a an = 0, f

f

′ ′a ′ ∇a ∇ φ

= = =

f

f

f

f

f

a ′ ′a ′ ′a ′n ′ ∇ a ∇ φ − C an ∇ φ = ∇ a ∇ φ

f

f

f

f

a a ∇a ∇ φ + ∇a ∇ ψ f

f

4 π ρ + ∇ a ∇ a ψ = 4 π ρ + ∇a ∇a ψ. f

(The penultimate equality holds because (∇, φ) satisfies (R3); and the argument for the final equality

is exactly the same as the one given for (4.2.28).) But ∇a ∇b ψ = 0 and, so, ∇a ∇b ψ = ta ξ n ∇n ∇b ψ,

where ξ n is any smooth, future-directed timelike field on O. It follows that ∇a ∇a ψ = 0 and, therefore, f

f

∇′a ∇′a φ′ = 4 π ρ, as required for (R4).

Just as with the Geometrization Lemma, only a small change is necessary here if we want to work with a cosmological constant. If we replace (4.2.18) with Rbc = 4 π ρ tb tc − Λ tb tc , then substitution for

Rbc in (4.2.27) yields ∇a φa + Λ = 4 π ρ. The further argument that φa is of the form ∇a φ is unaffected. So we are led to (4.2.11).

f

The Trautman Recovery Theorem tells us that if ∇ arises as the geometrization of the pair (∇, φ), f

then, for any field ψ such that ∇a ∇b ψ = 0, it also arises as the geometrization of (∇′ , φ′ ) where φ′ = φ + ψ f

f

and ∇′ = (∇, tb tc ∇a ψ).

f

(∇, φ)

f

(∇′ , φ′ )

ց ր

∇

240


f

f

We certainly have sufficient freedom here to insure that ∇′ is, in fact, distinct from ∇. We can think of

∇b ψ as the “spatial gradient” of ψ. The stated condition on ψ, namely ∇a ∇b ψ = 0, is just the requirement

that this spatial gradient be constant on spacelike hypersurfaces. The condition can certainly be satisfied without that gradient vanishing at all points. (Its value can change from one spacelike hypersurface to f

f

another.) And if ∇a ψ 6= 0 at some point p, then ∇′ cannot be the same operator as ∇. Indeed, let ξ a be

the four-velocity field of a timelike curve passing through p. Then at p, f

f

f

f

ξ a ∇′a ξ b = ξ a (∇a ξ b − (ta tn ∇b ψ)ξ n ) = ξ a ∇a ξ b − ∇b ψ 6= ξ a ∇a ξ b . We can use the current discussion to capture in precise language the standard claim that gravitational force in (standard) Newtonian theory is a gauge quantity. Consider a point particle with mass m and f

four-velocity ξ a that is not accelerating with respect to ∇. According to the de-geometrization (∇, φ), f

f

the particle has acceleration ξ n ∇n ξ a and is subject to gravitational force −m ∇ a φ = −m∇a φ. (We get

this from (R3).) Rather than being subject to no forces at all — the account given by the geometrized formulation of the theory — it is here taken to be subject to two “forces” (inertial and gravitational) f

that cancel each other. Alternatively, according to the de-geometrization (∇ ′ , φ′ ), it has acceleration f

f

f

ξ n ∇ ′n ξ a = ξ n ∇n ξ a − ∇a ψ and is subject to gravitational force −m ∇ ′a φ′ = −m∇a φ − m∇a ψ. So

the gravitational force on the particle is determined only up to a factor m∇a ψ, where ∇a ψ is constant

on any one spacelike hypersurface but can change over time.

Of course, if boundary conditions are brought into consideration, we regain the possibility of unique de-geometrization. In particular, if we are dealing with a bounded mass distribution, i.e., if ρ has compact support on every spacelike hypersurface, then it seems appropriate to require that the gravitational field die off as one approaches spatial infinity. But if ∇a ψ is constant on spacelike hypersurfaces and if it goes

to 0 at spatial infinity, then it must vanish everywhere.

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4.3

Interpreting the Curvature Conditions

In this section, we consider the geometric significance of three curvature conditions that appear in Trautman’s formulation of geometrized Newtonian gravitation theory: Rab

=

4πρ ta tb ,

(4.3.1)

Ra b c d

=

Rc d a b ,

(4.3.2)

Rab cd

=

0.

(4.3.3)

We start with the third. We know already (proposition 4.2.4) that it holds in a classical spacetime iff the latter is spatially flat (Rabcd = 0) and, at least locally, admits a unit timelike vector field ξ a that is rigid and twist-free (∇a ξ b = 0). We also have the following more direct interpretation. Proposition 4.3.1. Let (M, ta , hab , ∇) be a classical spacetime. Then Rab cd = 0 throughout M iff parallel transport of spacelike vectors within M is, at least locally, path independent.

Proof. (If) This direction is immediate. Let p be any point in M , and let O be an open set containing p within which parallel transport of spacelike vectors is path independent. We can certainly find three 1

2

3

i

smooth, linearly independent, spacelike fields σa , σa , σa on O that are constant (∇n σa = 0). (Start with three linearly independent, spacelike vectors at p and parallel transport them, along any curve, to other points in O.) For each one, we have Ra rcd σ r = −2 ∇[c ∇d] σ a = 0 i

1

2

i

3

at p. Since σa , σa , σa span the space of spacelike vectors at p, it follows that Ra rcd σr = 0 for all spacelike vectors σ a there. So Ra rcd hrb αb = 0 for all co-vectors αb at p, i.e., Rab cd = Ra rcd hrb = 0 at p. (Only if) There are various ways to see this. But it is, perhaps, easiest to make use of what we have established and reduce this to a claim about a (different) flat derivative operator. If Rab cd = 0, then, by proposition 4.2.4, given any point p in M , there is an open set O containing p and a future-directed unit timelike vector field η a on O such that ∇a η b = 0. Now recall our proof of the Trautman Recovery f

Theorem (proposition 4.2.5). Let φa be the acceleration field of η a , and let ∇ be the derivative operator f

on O defined by ∇ = (∇, C a bc ), where C a bc = tb tc φa . We established in our proof of the Recovery f

Theorem that ∇ is flat. (And for this part of the proof, we did not need the additional assumptions

that appear in our formulation of the theorem, namely Ra b c d = Rc d a b and Rbc = 4 π ρ tb tc . We needed f

only Rab cd = 0.) So parallel transport of all vectors within O relative to ∇ is, at least locally, path f

independent. To complete the proof, it suffices to note that ∇ and ∇ agree in their action on spacelike

vector fields (and so agree in their determinations of parallel transport for such fields on arbitrary curves). This is clear. For let σ a be a smooth spacelike vector field (defined on some open subset of O). Then f

b b b n b ∇a σ = ∇a σ − C an σ = ∇a σ ,

as required, since C b an σ n = (ta tn φb ) σ n = 0.

242


The proposition also provides a physical interpretation of the third curvature condition (4.3.3) in terms of the precession, or non-precession, of gyroscopes. Suppose we hold two spinning gyroscopes at a point, side by side, with their axes co-aligned. And suppose we then transport them (without constraint) to another point along different routes. We cannot expect a priori that, on arrival, their axes will still be co-aligned. There is no reason why “gyroscope propagation” must be path independent. Indeed, we see from the proposition that it will be path independent (at least locally) iff (4.3.3) holds. Now we consider the geometrized version of Poission’s equation Rab = 4 π ρ ta tb . The interpretation we offered for Einstein’s equation in terms of geodesic deviation has a close counterpart here. Almost everything carries over intact from section 2.7. Let ξ a be a “geodesic reference frame” defined on some open set in M , i.e., a smooth, future-directed, unit timelike vector field whose associated integral curves are geodesics. Further, let λa be a smooth, spacelike vector field along (the image of) one of the integral curves γ satisfying £ξ λa = 0. (Once again, we can think of λa as a connecting field that joins the image of γ to the image of an “infinitesimally close” neighboring integral curve.) The equation of geodesic deviation ξ n ∇n (ξ m ∇m λa ) = Ra bcd ξ b λc ξ d

(4.3.4)

carries over without alteration, as does the expression we derived for the “average radial acceleration” of ξa, 1 ARA = − Rbd ξ b ξ d . 3

(4.3.5)

The latter, in turn, leads to the following proposition (which is proved in almost exactly the same way as proposition 2.7.2). Proposition 4.3.2. Let (M, ta , hab , ∇) be a classical spacetime, let ρ be a smooth scalar field on M , and

let p be a point in M . Then Poisson’s equation Rab = 4 π ρ ta tb holds at p iff for all geodesic reference

frames ξ a (defined on some open set containing p), the average radial acceleration of ξ a at p is given by 4 ARA = − π ρ. 3 We can make the result look even more like proposition 2.7.2 if we use our alternate formulation of Poisson’s equation. In that case, the conclusion is this: Poisson’s equation Rab = 8 π (Tâb − 1 tab Tˆ) 2

holds at p iff for all geodesic reference frames ξ a (defined on some open set containing p), the average 8π 1 radial acceleration of ξ a at p is given by ARA = − π (Tâb − tab Tˆ) ξ a ξ b . 3 2 Finally, we turn to the geometric interpretation of the second condition in our list, Ra b c d = Rc d a b . This will require a good deal more work than the others. We show that it holds in a classical spacetime iff the latter admits, at least locally, a smooth, unit timelike field ξ a that is geodesic (ξ n ∇n ξ a = 0) and

twist-free (∇[a ξ b] = 0). This equivalence is proved in Dombrowski and Horneffer [13] and K¨ unzle [34]. Our argument, at least for the “only if” half (proposition 4.3.7), is a bit different from theirs. We begin with the “if” half of the assertion, which is straightforward.


243

Proposition 4.3.3. Let (M, ta , hab , ∇) be a classical spacetime, and let p be any point in M . Assume there is a smooth, future-directed, unit timelike field ξ a , defined on some open set containing p, that is

geodesic and twist-free. Then Rab cd = Rc dab at p. Proof. It suffices for us to show that, at p, contracting (Rab cd − Rc dab ) with (i) ξ b ξ d , (ii) hbr hds , and (iii)

ξ b hds (or hbr ξ d ) yields 0. The claim in case (ii) comes free, without any assumptions about ξ a , since Rarcs = Rcsar holds in any classical spacetime. (Recall (4.1.30).) For case (iii), we need only the fact that ξ a is twist-free. We must show that Rab cs ξ b = Rcsab ξ b . To do so, we recast the right side using symmetries of the curvature field, namely Ra[bcd] = 0, Rab(cd) = 0, and R

(ab) cd

= 0. (The first two hold for any derivative operator. The third follows from the compatibilty of ∇

with hab . Recall (4.1.26). We use the symmetries with some indices in raised position. So, for example, since Rabcd + Radbc + Racdb = 0, it follows that Rab cd + Radb c + Racdb = 0.) Rcsab ξ b

=

−Rscab ξ b = Rsb ca ξ b + Rsab c ξ b = Rsb ca ξ b − Rasb c ξ b

=

Rsb ca ξ b + (Racsb + Rab cs ) ξ b = Rsb ca ξ b + Rab cs ξ b − Rcasb ξ b

=

Rsb ca ξ b + Rab cs ξ b + (Rcb as + Rcs ba ) ξ b .

Hence, Rcsab ξ b

1 s ca b (R b ξ + Rab cs ξ b + Rcb as ) ξ b 2 = −(∇[c ∇a] ξ s + ∇[c ∇s] ξ a + ∇[a ∇s] ξ c ). =

If we now expand the right side and use the fact (for the first time) that ∇a ξ b = ∇b ξ a , we arrive at Rcsab ξ b = −(∇c ∇s ξ a − ∇s ∇c ξ a ) = Rab cs ξ b . Finally, we consider case (i). Here we need both the fact that ξ a is twist-free and that it is geodesic. We must show that Rab cd ξ b ξ d = Rc dab ξ b ξ d , i.e., that Rab cd ξ b ξ d is symmetric in a and c. But Rab cd ξ b ξ d

= −ξ d (∇c ∇d ξ a − ∇d ∇c ξ a ) = −∇c (ξ d ∇d ξ a ) + (∇c ξ d )(∇d ξ a ) + ξ d ∇d ∇c ξ a .

The first term on the far right vanishes since ξ a is geodesic. The third is symmetric in a and c since ξ a is twist-free. The second is symmetric in a and c for the same reason, since (∇c ξ d )(∇d ξ a ) = (∇d ξ c )(∇d ξ a ) = hdn (∇n ξ c )(∇d ξ a ) = (∇n ξ c )(∇n ξ a ) = (∇n ξ c )(∇a ξ n ). Next we consider a particular class of derivative operators that satisfy the curvature condition (4.3.2) (in addition to being compatible with the background metrics ta and hab ). Proposition 4.3.4. Let (M, ta , hab , ∇) be a classical spacetime, and let ξ a be any smooth, unit, futuree on M such that (i) directed timelike vector field on M . Then there exists a unique derivative operator ∇

e is compatible with ta and hab and (ii) ξ a is geodesic and twist-free with respect to ∇. e ∇


244

e the special derivative operator determined by ξ a . It When conditions (i) and (ii) obtain, we call ∇

follows immediately from the preceding proposition that all special derivative operators (determined by some field) satisfy (4.3.2). We shall soon verify (in proposition 4.3.7) that they are the only derivative operators that do so. ˆ ab be the projection field associated with ξ a , let κab = h ˆ n[b ∇a] ξ n , let C a bc = 2 t(b κc) a and, Proof. Let h e = (∇, C a bc ). Then, by proposition 4.1.3 , ∇ e is compatible with ta and hab . Moreover, we finally, let ∇ ˆ ab hbc = δa b −ta ξ b , e To see this, note first that since h claim, ξ a is geodesic and twist-free with respect to ∇. we have

κa b = hbr κar =

1 br ˆ ˆ na ∇r ξ n ) = 1 (∇a ξ b − h ˆ na ∇b ξ n ) h (hnr ∇a ξ n − h 2 2

and, therefore,

Now

κab

=

κa b ξ a

=

1 a b (∇ ξ − ∇b ξ a ) = ∇[a ξ b] , 2 1 a ξ ∇a ξ b . 2

e a ξ b = ∇a ξ b − C b ar ξ r = ∇a ξ b − (ta κr b + tr κa b )ξ r . ∇

Hence, since κab is anti-symmetric, we have e [a ξ b] ∇

e a ξb ξa∇

= ∇[a ξ b] − κab = 0, = ξ a ∇a ξ b − 2 κa b ξ a = 0,

as claimed. So we have established existence. e e = (∇, e C e a ) is a second derivative operator on M that satisfies conditions For uniqueness, suppose ∇ bc e e are compatible with ta and hab ) e and ∇ (i) and (ii). We know from proposition 4.1.3 (since both ∇

ea = 2 han t(b κc)n . We show that κcn = 0. that there is a smooth, anti-symmetric field κab such that C bc

Now κcn ξ c ξ n = 0, since κcn is anti-symmetric. So it will suffice for us to show that κcn ξ c hns = 0 and e e we have, first, e and ∇, κcn hcr hns = 0. Since ξ a is geodesic with respect to both ∇ Next,

e e c ξ a = ξ c (∇ e c ξa − C e a ξ b ) = −C e a ξ b ξ c = −2 κcn ξ c han . 0 = ξc∇ bc bc e e e c ξ s = hrc (∇ e c ξs − C es ξ b ) = ∇ e r ξ s − κrs . e r ξ s = hrc ∇ ∇ bc

e e [r ξ s] = 0 = ∇ e [r ξ s] (and since κcn is anti-symmetric), we also have hrc hsn κcn = 0. So, since ∇

Now we extend proposition 4.1.3 and consider the most general form for a connecting field C abc that

links two derivative operators on M that are compatible with ta and hab and also satisfy (4.3.2).


245

Proposition 4.3.5. Let (M, ta , hab , ∇) be a classical spacetime such that Ra b c d = Rc d a b . Let ∇′ =

(∇, C abc ) be a second derivative operator on M where C abc = 2 han t(b κc)n and κab is a smooth, antisymmetric field on M . (We know this is the general form for a derivative operator on M that is compatible with ta and hab .) Then R′a b c d = R′c d a b iff κab is closed, i.e., ∇[n κab] = 0. (Here, of course, R′a bcd is the Riemann curvature field associated with ∇′ .)

Proof. We know (from problem 1.8.1) that R′abcd = Rabcd + 2 ∇[c C ad]b + 2 C nb[c C ad]n . In the present case, where C abc = 2t(b κc) a , we have C nbc C adn = td tb κc n κn a + td tc κb n κn a and, hence 2 C nb[c C ad]n = 2 tb t[d κc] n κn a . Similarly, ∇c C adb = td ∇c κb a + tb ∇c κd a and, hence, 2 ∇[c C ad]b = 2 t[d ∇c] κb a + 2 tb ∇[c κd] a . If we now raise the index c in all these terms, we arrive at R′a b c d = Ra b c d + (td ∇c κb a + tb ∇c κd a − tb ∇d κca ) + tb td κcn κn a and, therefore, also R′c d a b = Rc d a b + (tb ∇a κd c + td ∇a κb c − td ∇b κac ) + td tb κan κn c . We are assuming that Ra b c d = Rc d a b . And, by the anti-symmetry of κab , κcn κn a = κan κn c . So we see that R′a b c d = R′c d a b iff the respective middle terms (in parentheses) in the two lines are equal, i.e., iff tb (−∇a κd c + ∇c κd a − ∇d κca ) = td (−∇c κb a + ∇a κb c − ∇b κac ).

(4.3.6)

In turn, this equation holds iff −∇a κd c + ∇c κd a − ∇d κca = 0.

(4.3.7)

(Why? If (4.3.7) holds, then both sides of (4.3.6) vanish. Conversely, assume (4.3.6) holds, let ψ ac d = (−∇a κd c + ∇c κd a − ∇d κca ), and let ξ a be any unit timelike vector field. Contracting both sides of

(4.3.6) with ξ b hdr yields hdr ψ ac d = 0. Contracting both sides with ξ b ξ d yields ξ d ψ ac d = 0. So it must be the case that ψ ac d = 0.) We can express (4.3.7) in the form har hcs ∇[r κsd] = 0.

(4.3.8)

∇[r κsd] = 0.

(4.3.9)

But this condition is equivalent to

For if (4.3.8) holds, then, by the anti-symmetry of ∇[r κsd] , contraction with ξ r ξ s ξ d , ξ r ξ s hdn , ξ r hsc hdn , and hra hsc hdn all yield 0. Thus, as claimed, R′a b c d = R′c d a b iff κab is closed.

246


Now we make precise a sense in which condition Rab cd = Rcdab rules out the possibility of “spontaneous rotation”. Proposition 4.3.6. Let (M, ta , hab , ∇) be a classical spacetime such that Rab cd = Rc da b . Let ξ a be a smooth, future-directed, unit timelike field on M that is geodesic (with respect to ∇). Then its associated rotation and expansion fields satisfy

ξ n ∇n ω ab = 2 ω n[a θn b] .

(4.3.10)

Hence, given any integral curve γ : I → M of ξ a , if ξ a is twist-free (ω ab = 0) at one point on γ[I], it is twist-free at all points on it. (Or, more colloquially, if it is twist-free at one time, it is twist-free at all times.) Proof. We know that ∇[a ξ b] = ω ab . (This follows immediately from (4.1.43).) Hence, 2 ξ n ∇n ω ab

Since ξ a is geodesic,

= ξ n ham ∇n ∇m ξ b − ξ n hbm ∇n ∇m ξ a = ξ n ham ∇m ∇n ξ b − Rbsnm ξ s − ξ n hbm (∇m ∇n ξ a − Rasnm ξ s )

ξ n ham ∇m ∇n ξ b = ham ∇m (ξ n ∇n ξ b ) − (∇m ξ n )(∇n ξ b ) = −(∇a ξ n )(∇n ξ b )

and, similarly, −ξ n hbm ∇m ∇n ξ a = (∇b ξ n )(∇n ξ a ). Furthermore, since Rb s(nm) = 0,

−ξ n ham Rbsnm ξ s + ξ n hbm Rasnm ξ s = Rb s a n − Ra s b n ξ n ξ s = Rb s a n − Ra n b s ξ n ξ s = 0.

So,

2 ξ n ∇n ω ab

= −(∇a ξ n )(∇n ξ b ) + (∇b ξ n )(∇n ξ a ) = −[2 ∇[a ξ n] + ∇n ξ a ](∇n ξ b ) + [2 ∇[b ξ n] + ∇n ξ b ](∇n ξ a ) = −2 (∇[a ξ n] )(∇n ξ b ) + 2 (∇[b ξ n] )(∇n ξ a ) = −2 ω an (θn b + ωn b ) + 2 ω bn (θn a + ωn a ) = −2 ω an θn b + 2 ω bn θn a = 4 ω n[a θn b] .

Now let γ : I → M be an integral curve of ξ a , and suppose ωab = 0 at some point γ(s0 ). It follows

from the basic uniqueness theorem for systems of first-order ordinary differential equations that (4.3.10) will be satisfied at all points on γ[I] iff ωab = 0 vanishes everywhere on that set. (To see this in detail, 1

4

let σ a , ..., σ a be a basis for the co-tangent space at some point on γ[I] that is orthonormal with respect to hab (in our extended sense of “orthonormal”). We can extend the vectors (by parallel transport) to i

i

fields σ a on γ[I] — we use the same notation — that satisfy ξ n ∇n σ a = 0. Since ∇ is compatible with

hab , the generated fields will be orthonormal everywhere. Now consider the scalar (coefficient) fields ij

i

i

ω = ω ab σ a σ b . Equation (4.3.10) can then be expressed as a system of first-order differential equations ij

dω 11 12 44 = fij ( ω , ω , ..., ω ) dt to which the uniqueness theorem is applicable.)

247


We have claimed that condition (4.3.2) holds iff, at least locally (in a neighborhood of every point), there exists a unit timelike vector field that is geodesic and twist-free. We have proved the “if” half of the claim (in proposition 4.3.3). Now, finally, we turn to the converse. Proposition 4.3.7. Let (M, ta , hab , ∇) be a classical spacetime such that Rab cd = Rc dab . Then, given

any point p in M , there is a smooth, future-directed, unit timelike vector field, defined on some open set containing p, that is geodesic and twist-free (with respect to ∇). Proof. Let p be given. Our proof will proceed in two steps and make reference to three smooth, futuredirected, unit timelike fields: ξ a , ξ ′a , and ξ ′′a . (They will be defined on open sets O, O′ , and O′′ , respectively, where p ∈ O′′ ⊆ O′ ⊆ O.) ξ a will be an arbitrary field. ξ ′a will be twist-free. ξ ′′a will be

geodesic and twist-free. (It is the existence of the third that we need to establish.)

(Step 1) Let ξ a be a smooth, future-directed, unit timelike field defined on some open set O containing e on O such that ∇ e is compatible with ta and p. By proposition 4.3.4, there is a derivative operator ∇ e Let C a be the connecting field (on hab , and such that ξ a is geodesic and twist-free with respect to ∇. bc

e C a ). Now, by proposition 4.3.3, R ea c = R ec a . So, since since both ∇ and ∇ e O) such that ∇ = (∇, bc b d d b

satisfy (4.3.2), it follows by proposition 4.3.5 that there is a smooth, closed, anti-symmetric field κab on

O such that C abc = 2 han t(b κc)n . Since κab is closed, we know by proposition 1.8.3 that it is, at least locally, exact. So there is an open subset O′ of O containing p, and a smooth field κa on O′ such that e [a κb] . Now consider the field ξ ′a = ξ a + κa on O′ . It is a smooth, future-directed, unit timelike κab = ∇

field. (It is of unit timelike length since ta κa = ta hab κb = 0.) We claim that it is twist-free with respect to ∇. We have

But,

e n (ξ a + κa ) − C amn (ξ m + κm ). ∇n ξ ′a = ∇n (ξ a + κa ) = ∇

C amn (ξ m + κm ) = (tm κn a + tn κm a )(ξ m + κm ) = κn a + tn κm a (ξ m + κm ). Hence, e nξa + ∇ e n κa − κna ∇n ξ ′a = ∇

e and κab = ∇ e [a κb] ), and, therefore (since ξ a is twist-free with respect to ∇, e [n ξ a] + ∇ e [n κa] − κna = 0. ∇[n ξ ′a] = ∇

(Step 2) So far, we established the existence of a field ξ ′a , defined on some open set O′ containing p, that is twist-free with respect to ∇. Now let S be a spacelike hypersurface within O′ that contains p.

Then we can find a smooth, future-directed, unit timelike vector field ξ ′′a , defined on some open subset O′′ of O′ containing p, that is geodesic with respect to ∇ and agrees with ξ ′a on S. (We first restrict

ξ ′a to S, and then use each vector in this restricted field ξ ′a |S to generate a geodesic. This gives us a

congruence of curves. We take ξ ′′a to be its tangent field.) Now, since ξ ′a is twist-free on S, so is ξ ′′a . (The difference field (ξ ′′a − ξ ′a ) vanishes on S. So, at any point of S, its directional derivative in any


248

spacelike direction vanishes as well, i.e., han ∇n (ξ ′′b − ξ ′b ) = 0. Hence, on S, ∇[a ξ ′′b] = ∇[a ξ ′b] = 0.) But

now, since the geodesic field ξ ′′a is twist-free on S, it follows from proposition 4.3.6 that is it everywhere twist-free. So we are done.

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4.4

A Solution to an Old Problem about Newtonian Cosmology

The geometrized formulation of Newtonian theory provides a satisfying solution to an old problem about Newtonian cosmology. We present it in this section.3 At issue is whether Newtonian gravitation theory provides a sensible prescription for what the gravitational field should be like in a hypothetically infinite, homogeneous universe. Let us first think about this in terms of a traditional, non-geometrized, three-dimensional formulation of the theory. Let (R3 , gab ) be three-dimensional Euclidean space. We take it to represent physical space at a given time. Further, let ρ and φ be two smooth functions on R3 that, respectively, give the mass density and the gravitational potential at different points of space.4 We assume that they satisfy Poisson’s equation ∇a ∇a φ = 4 π ρ.

(Here ∇ is the derivative operator on R3 compatible with gab .)

Suppose that we are dealing with a homogeneous distribution of matter, i.e., suppose that ρ is constant. Then, presumably, the gravitational field associated with this matter distribution should be homogeneous as well. (Why should it be different here from the way it is there?) The gravitational force felt by a particle of unit mass at any point is given by −∇a φ. So, it would seem, the natural way to capture the

homogeneity condition on the gravitational field is to require that the field ∇a φ be constant, i.e., require

that ∇b ∇a φ = 0. But now we have a problem. If ∇b ∇a φ = 0, and if Poisson’s equation is satisfied, then

4 π ρ = ∇a ∇a φ = 0. So we cannot satisfy the homogeneity condition except in the degenerate case where

the mass density ρ is everywhere 0.

Here is another version of the problem. It directs attention to a particular class of solutions to Poisson’s equation ∇a ∇a φ = 4 π ρ that do exist in the case where ρ is constant (∇a ρ = 0). Let o be any point in

R3 , and let χa be the position field determined relative to o. So ∇a χb = δa b , and χa = 0 at o. Let us say that a smooth field φ on R3 is a canonical solution centered at o if ∇a φ =

4 π ρ χa , 3

(4.4.1)

i.e., if ∇a φ is a spherically symmetric, outward-directed, radial vector field, centered at o, whose assign-

ment to any point p has length 43 π ρ r, where r is the Euclidean distance between o and p. Note that if this condition holds, then (since ∇a ρ = 0), ∇a ∇a φ =

4 π ρ (∇a χa ) = 4 π ρ. 3

So canonical solutions centered at o (if they exist) are solutions. And they certainly do exist, e.g., φ = 3 For

2 π ρ (χn χn ). 3

further discussion of the problem and its history, see Norton [43, 44, 45] and Malament [40].

4 Caution:

we have previously understood ρ and φ to be objects defined on a four-dimensional spacetime manifold, and

shall soon do so again. But now, temporarily, we take them to be defined on a three-dimensional manifold (representing space a given time) instead.


250

Not all solutions to Poisson’s equation (in the present case where ρ is constant) are canonical solutions centered at some point or other. (If φ is a solution, then so is (φ + ψ), where ψ is any smooth field that satisfies ∇a ∇a ψ = 0.) But canonical solutions are the only solutions that satisfy a certain natural constraint, and for this reason they are the only ones that are usually considered in discussions of Newtonian cosmology. The constraint arises if we consider not just the distribution of cosmic matter at a given time, but also its motion under the influence of that potential. It turns out that if we require that the motion be isotropic in a certain natural sense, then all solutions are ruled out except those that are canonical for some center point o. (We shall, in effect, prove this. See proposition 4.4.3.) In any case, our problem re-emerges when we direct our attention to the class of canonical solutions. The gravitational field associated with any one of them is a radial field that vanishes at a unique center point. Why, one wants to ask, should there be any such distinguished point in a homogeneous universe? And why should it be one point rather than another, i.e., why should any one canonical solution be a better choice for the gravitational field in a homogeneous universe than another? That is the problem. A solution, or dissolution, can be found in the recognition that the gravitational field (in standard formulations of Newtonian theory) is a kind of “gauge field”, i.e., a field that is, in general, systematically underdetermined by all experimental evidence. Despite appearances, canonical solutions centered at different points really are empirically equivalent. No experimental test could ever distinguish one from another (or distinguish the center point of any one of them). Canonical solutions centered at different points should be viewed as but alternative mathematical representations of the same underlying state of gravitational affairs — a state that is perfectly homogenous in the appropriate sense. One can certainly argue for these claims directly, without reference to geometrized formulations of Newtonian theory.5 (See, for example, Heckmann and Sch¨ ucking [31] and Norton [44].) But some insight is achieved if we do think about this old problem in Newtonian cosmology using the ideas developed in section 4.2. We can develop an account of Friedmann-like cosmological models within geometrized Newtonian gravitation theory, and then recover the class of canonical solutions (centered at different points) as but alternative “de-geometrizations” of the initial curved derivative operator — exactly as described at the end of that section. The choice between different canonical solutions emerges as a choice between different ways to decompose into “gravitational” and “inertial” components the net force experienced by a point particle. Nothing more. Before proceeding, we give an alternative characterization of the class of canonical solutions — at least in the case of interest where ρ > 0 — that will be convenient later. 5 The

important point is that if φ and φ′ are canonical solutions, based at o and o′ , respectively, the difference field

(∇a φ − ∇a φ′ ) is constant, and constant gravitational fields are undetectable. Only field differences can be detected. The

difference field is constant since

4 4 4 ∇a (∇b φ − ∇b φ′ ) = ∇a ( π ρ χb − π ρ χ′b ) = π ρ (δa b − δa b ) = 0. 3 3 3

251


Proposition 4.4.1. Let (R3 , gab ) be three-dimensional Euclidean space, and let ρ be a constant field on R3 with ρ > 0. Then for all smooth fields φ on R3 , the following conditions are equivalent. (1) φ is a canonical solution (to Poisson’s equation ∇a ∇a φ = 4 π ρ) centered at some point in R3 . 4 (2) ∇a ∇b φ = π ρ g ab . 3 Proof. One direction is immediate. If φ is a canonical solution centered at point o (and if χa is the position field relative to o), 4 4 4 ∇a ∇b φ = ∇a ( π ρ χb ) = π ρ (∇a χb ) = π ρ g ab . 3 3 3 Conversely, suppose φ satisfies condition (2). Let φ′ be a canonical solution centered at some point o′ , let χ′a be the position field relative to o′ , and let κb be the difference field κb = ∇b φ − ∇b φ′ = ∇b φ −

4 π ρ χ′b . 3

Then κb is constant (∇a κb = 0) and ∇b φ =

4 πρ 3

3 χ′b + ( ) κb . 4πρ

Now let o be the (unique) point where the vector field on the right side vanishes. (We can think of o as the point one gets if one displaces o′ by the vector −(3/4 π ρ) κb . This makes sense since we can identify

vectors at different points in three-dimensional Euclidean space.) Then (χ′b + (3/4 π ρ) κb ) is just what we would otherwise describe as the position field χb relative to o. (Note that when we apply ∇a to the

field we get δa b .) So φ qualifies as a canonical solution centered at o.

Note that the proposition fails if ρ = 0. In that case, the implication (1) ⇒ (2) still holds, but not the

converse. For then all canonical solutions have vanishing gradient (∇a φ = (4/3) π ρ χa = 0), whereas condition (2) requires only that ∇a φ be constant. Condition (2) in the proposition naturally lifts to the context of classical spacetimes where it becomes ∇a ∇b φ =

4 π ρ hab . 3

(4.4.2)

(That is why it will be convenient later.) The latter holds iff the restriction of φ to any spacelike hypersurface S (together with the restrictions of ∇ and hab to S) satisfies (2). Let us now shift back to the framework of geometrized Newtonian gravitation theory. Our first task is to introduce a class of cosmological models that correspond to the Friedmann spacetimes we considered in section 2.11. We could proceed just as we did there, i.e., start with a condition of spatial homogeneity and isotropy (relative to some smooth, future-directed, unit timelike field ξ a ) and derive the consequences of that assumption. We could show again that ξ a is necessarily geodesic, twist-free, and shear-free; that any vector field definable in terms of the basic elements of structure ta , hab , ∇, and ξ a is necessarily

proportional to ξ a ; and so forth. Instead, we proceed directly to an explicit characterization.

252


Let us first take a (classical) cosmological model to be a a structure of the form (M, ta , hab , ∇, ξ a , ρ),

where (M, ta , hab , ∇) is a classical spacetime, ξ a is a smooth, future-directed unit timelike field on M ,

and ρ is a smooth field on M . We take ξ a be the four-velocity of a cosmic fluid that fills all of spacetime, and take ρ to be the mass-density of the fluid. Next, let us say that (M, ta , hab , ∇, ξ a , ρ) is Friedmann-like

if the following conditions are satisfied.

(1) ξ a is geodesic, twist-free, and shear-free, i.e., ∇a ξ b =

1 b (δa − ta ξ b ) θ. 3

(4.4.3)

(Here θ = ∇a ξ a is the scalar expansion field associated with ξ a . Note that (4.4.3) follows from

(4.1.42), (4.1.45), and (4.1.12). In more detail, since θab = σab = 0 and ξ n ∇n ξ a = 0, we have 1 1 ˆ ∇a ξ b = θa b = θan hnb = θ han hnb = (δa b − ta ξ b ) θ.) 3 3 (2) ∇a ρ = 0, i.e., ρ is constant on all spacelike hypersurfaces. (3) Poisson’s equation Rab = 4 π ρ tab holds.

Note that we have not included Trautman’s two supplemental integrability conditions (Rab cd = Rc dab and Rab cd = 0) in the list. We have not done so because, as we now show, they follow from the other assumptions. So in this special case — the case of Friedmann-like cosmological models — the difference between our two formulations of geometrized Newtonian theory collapses. (We shall consider another case where it collapses in section 4.5.) Proposition 4.4.2. Let (M, ta , hab , ∇, ξ a , ρ) be a Friedmann-like cosmological model. Then the following

conditions hold.

(1) Rab cd = Rc da b and Rab cd = 0. (2) ξ n ∇n θ = −4 π ρ −

1 3

θ2 .

Proof. (1) The first condition Ra b c d = Rc d a b follows immediately from proposition 4.3.3. (We need only that ξ a be geodesic and twist-free for this much.) For the second condition, Rab cd = 0, it will suffice to establish the existence, at least locally, of a smooth, future-directed, unit timelike field η a on M that is rigid and twist-free (∇a η b = 0). For then we can invoke proposition 4.2.4. Let p be any point in M . All Friedmann-like cosmological models are spatially flat (by proposition 4.1.5). So there must be an open set O containing p and a smooth spacelike field χa on O such that ∇a χb = hab . (Recall the very beginning of our proof of proposition 4.2.3.) Now consider the field ηa = ξ a −

1 θ χa 3

on O. It is certainly a smooth, future-directed, unit timelike field. We claim that it is rigid and twist-free, as required. To see this, note first that, by (4.4.3), ∇a η b = ∇a ξ b −

1 1 1 1 1 θ ∇a χb − (∇a θ) χb = ( θ hab − θ hab ) − (∇a θ) χb . 3 3 3 3 3

253


Furthermore, ∇a θ = 0. This follows, since by (4.4.3) and Poisson’s equation, ∇a θ

= =

han ∇n ∇m ξ m = −han Rm rnm ξ r + han ∇m ∇n ξ m

1 1 −han Rrn ξ r + ∇m (∇a ξ m ) = −han (4 π ρ trn ) ξ r + ∇m ( ham θ) = ∇a θ. 3 3

So ∇a η b = 0, as claimed. (2) Here we start as we did in our derivation of Raychaudhuri’s equation (2.8.17): ξ a ∇a θ

=

ξ a ∇a ∇b ξ b = −ξ a Rbcab ξ c + ξ a ∇b ∇a ξ b

=

−Rca ξ c ξ a + ∇b (ξ a ∇a ξ b ) − (∇b ξ a )(∇a ξ b ).

1 2 θ . And by Poisson’s equation (the third 3 condition in our characterization of Friedmann-like cosmological models), Rca ξ c ξ a = 4 π ρ. So we are

But now, by (4.4.3), ξ a ∇a ξ b = 0 and (∇b ξ a )(∇a ξ b ) = done.

Note that condition (2) in the proposition — the equation that governs the rate of change of θ in Friedmann-like cosmological models — agrees with (2.11.9) in the case where p = 0. This makes sense. Though in general relativity the “gravitational field” generated by a blob of perfect fluid depends on its internal pressure as well as on its mass density, only the latter plays a role in Newtonian gravitation theory. Now we make precise our claim about the recovery of canonical solutions. Condition (4.4.4) in the following proposition is the condition we motivated using proposition 4.4.1. At least if ρ 6= 0, we can

understand it to capture the claim that the restriction of φ to any spacelike hypersurface is a canonical solution to Poisson’s equation. (If ρ = 0, it asserts instead that ∇a φ is constant on spacelike hypersur-

faces.)

Proposition 4.4.3. Let (M, ta , hab , ∇, ξ a , ρ) be a Friedmann-like cosmological model, and let φ be a f

smooth field on some open set in M . If φ arises as part of a de-geometrization (∇, φ) of ∇ (on that open

set), then

∇a ∇b φ =

4 π ρ hab . 3

(4.4.4) f

Conversely, if φ satisfies (4.4.4), then, at least locally, there is a derivative operator ∇ on M such that f

f

(∇, φ) is a de-geometrization of ∇. (Once again, to say that (∇, φ) is a de-geometrization of ∇ is to say

that it satisfies conditions (R1) - (R4) in the Trautman Recovery Theorem.)

Proof. We begin the proof of showing that, given any point p in M , there is an open set O containing p f

and some de-geometrization (∇ ∗ , φ∗ ) of ∇ on O such that ∇a ∇b φ∗ =

4 π ρ hab . 3

(4.4.5)

This will require a bit of work. But once we have established this much, our principal claims will follow easily.

254


We begin just as we did in our proof of the Trautman Recovery Theorem. (Note that all the assumptions needed for the theorem hold. In particular, the supplemental integrability conditions Ra b c d = Rc d a b and Rab cd = 0 hold. We know this from proposition 4.4.2.) Let p be any point in M . Then we can find an open set O containing p, and a smooth, future-directed, unit timelike field η a on O that is f

f

rigid and twist-free. Now consider the derivative operator ∇∗ on O defined by ∇∗ = (∇, C a bc ), where C a bc = ta tb φa and φa = ξ n ∇n ξ a . As we know from our proof of the Trautman Recovery Theorem, we

can (after possibly restricting O to some smaller open set containing p) find a smooth scalar field φ∗ on f

O such that φa = ∇a φ∗ and such that (∇ ∗ , φ∗ ) qualifies as a de-geometrization of ∇ on O. We claim

that φ∗ satisfies (4.4.5).

To see this, consider the field ξ a . (It gives the four-velocity of matter in our Friedmann-like cosmological model.) It is a geodesic field with respect to ∇. So, by condition (R3) in the Trautman Recovery Theorem, f

f

ξ n ∇∗n ξ a = − ∇∗a φ∗ . Hence, f

∗a

∇

f

∗b ∗ ∇ φ

f

f

f

f

f

f

f

f

f

− ∇∗a (ξ n ∇∗n ξ b ) = −(∇∗a ξ n )(∇∗n ξ b ) − ξ n ∇∗a ∇∗n ξ b

=

f

= −(∇∗a ξ n )(∇∗n ξ b ) − ξ n ∇∗n ∇∗a ξ b .

f

(4.4.6) f

(We use the fact that ∇∗ is flat for the final equality.) Next, we derive an expression for ∇∗n ξ b . We have,

by (4.4.3),

f

∗

∇nξ

b

= ∇n ξ b − C b nm ξ m = ∇n ξ b − (tn tm φb )ξ m = ∇n ξ b − tn φb 1 b (δn − tn ξ b ) θ − tn φb . = 3

It follows that

f

∗a b

ξ

∇ f

=

1 ab h θ. 3

f

Substituting these expression for ∇∗n ξ b and ∇∗a ξ b in (4.4.6) yields f

∗a

∇

f 1 ab 2 1 ab n f ∗ ∗b ∗ ∇ φ = − h θ − h ξ ∇ n θ. 9 3

Now by the second clause of proposition 4.4.2, f 1 ξ n ∇∗n θ = ξ n ∇n θ = − θ2 − 4 π ρ. 3 f

So, after substituting this expression for ξ n ∇∗n θ in (4.4.7), we have f

∗a

∇

f

∗b ∗ ∇ φ =

4 π ρ hab . 3

But f

∗a

∇

f

∗b ∗ ∇ φ

= =

f

∗a b ∗ am b ∗ b n ∗ ∇ ∇ φ = h (∇m ∇ φ − C mn ∇ φ )

ham [∇m ∇b φ∗ − (tm tn φb ) ∇n φ∗ ] = ∇a ∇b φ∗ .

(4.4.7)

255


So ∇a ∇b φ∗ =

4 π ρ hab , 3

as claimed. This completes the first part of the proof. Now let φ be a smooth field on some open subset U of M . Let p be any point in U . We know from what we have just proved that we can find an open subset O of U containing p and a de-geometrization f

(∇ ∗ , φ∗ ) of ∇ on O such that φ∗ satisfies (4.4.5). Suppose first that φ arises as part of a de-geometrization f

f

f

(∇, φ) of ∇ on U . Then we have two de-geometrizations of ∇ on O, namely (∇ ∗ , φ∗ ) and (∇, φ). By

the final part of the Trautman Recovery Theorem governing the non-uniqueness of de-geometrizations, it follows that 0 = ∇a ∇b (φ − φ∗ ) = ∇a ∇b φ − f

4 π ρ hab . 3 f

f

f

(Here the roles of (∇′ , φ′ ) and (∇, φ) in that theorem are played, respectively, by (∇, φ) and (∇ ∗ , φ∗ ).)

So φ satisfies (4.4.4) throughout the open set O containing p. But p was chosen arbitrarily. So φ satisfies (4.4.4) everywhere in U . Conversely, suppose φ satisfies (4.4.4). Then, by (4.4.5) again, we have ∇a ∇b (φ − φ∗ ) =

4 4 π ρ hab − π ρ hab = 0 3 3 f

f

on O. Hence, by the final part of the Trautman theorem again, if we set ∇ = (∇ ∗ , tb tc ∇a (φ − φ∗ )), f

then (∇, φ) qualifies as a de-geometrization of ∇ on O.

Let us think about what we would experience if we resided in a Friedmann-like Newtonian universe of the sort we have been considering. Suppose we were at rest in the cosmic fluid, i.e., moving along an integral curve of the background four-velocity field ξ a . Then we would experience no net force and would observe all other mass points in the fluid moving uniformly away from, or toward, us. If we were inclined to describe the situation in terms of traditional, non-geometrized Newtonian theory, we would say (adopting, implicitly, a particular de-geometrization) that the the gravitational field is centered where we are and vanishes there. (That is why we experience no net force.) But we would offer a different account for why our colleagues co-moving with other cosmic mass points experience no net force. From our point of view (i.e., according to our de-geometrization), they do experience a non-zero gravitational force. But it is perfectly balanced by a corresponding “inertial” force. And it is for this reason that they experience no net force. (Of course, those colleagues have their own story to tell with the roles reversed. They take themselves to be the ones residing where the gravitational field vanishes.)


4.5

256

Geometrized Newtonian Theory — Second Version

In this section, we prove a recovery or de-geometrization theorem for the K¨ unzle-Ehlers version of geometrized Newtonian gravitation theory. It is the counterpart to the recovery theorem we proved for the Trautman version (proposition 4.2.5) and actually subsumes that earlier result as a special case. We also consider a second set of special circumstances in which the difference between our two versions of the theory collapses. We saw in section 4.4 that Trautman’s second integrability condition Rab cd = 0 holds automatically in Friedmann-like cosmological models. Here we show that it holds automatically if we restrict attention to classical spacetimes that are, in a certain weak sense, asymptotically flat. We start with a lemma. Our proof of the Trautman Recovery Theorem turned on the availability of a rigid, twist-free field η a . Existence was guaranteed by the second integrability condition (proposition 4.2.4). Now we have to work with less. We cannot count on the existence of rigid, twist-free fields. But, as we now show, we can still count on the existence of fields that are, at least, rigid. And this will suffice. To prove the new recovery theorem, we need only rerun the argument for the old one using a field η a that is merely rigid. The computations are a bit more complicated, but no new ideas are involved. (We could have proved this version of the theorem first and then recovered the Trautman version simply by considering what happens when η a is also twist-free. But there is some advantage to taking on complications one at a time.) Proposition 4.5.1. Let (M, ta , hab , ∇) be a classical spacetime that is spatially flat (Rabcd = 0). Then,

given any point p in M , there exist an open set O containing p and a smooth, future-directed, unit timelike field η a on O that is rigid (∇(a η b) = 0).

Proof. Let p be any point in M , and let γ : I → M be a smooth, future-directed, timelike curve, with

four-velocity field ηba , that passes through p. We claim first that we can find three smooth, linearly 1

2

3

independent, spacelike fields σa , σa , σa on some open set O containing p with these properties (for all i): 1

1

2

2

3

3

(i) hab = σ a σ b + σ a σ b + σ a σ b . i

(ii) ∇a σ b = 0. i

(iii) ηbn ∇n σ a = 0 on γ[I].

We can generate the fields as follows. First we find three linearly independent, spacelike vectors at p that satisfy condition (i) — just as we did in the proof of proposition 4.1.4. Then we extend the vectors to an open set containing p in two stages. First, we extend them by parallel transport along γ. (So condition (iii) is satisfied.) Then we extend them “outward” from γ[I] by parallel transport along spacelike curves. The latter operation works this way. Let S be a spacelike hypersurface that intersects the image of γ at the point q. Then, because of spatial flatness, parallel transport of spacelike vectors within S is, at least locally, path independent. (Recall proposition 4.1.5.) So we can unambiguously extend the triple

257


1

2

3

σ a , σ a , σ a at q by parallel transport to points on S sufficiently close to q. The fields generated by this i

construction are “constant in spacelike directions”, i.e., λn ∇n σ a = 0 for all spacelike vectors λa . The

latter condition is equivalent to (ii). Finally, we claim, condition (i) holds everywhere. Consider the 1

1

2

2

3

3

difference field (hab − (σ a σ b + σ a σ b + σ a σ b )). It vanishes at p. Hence, by (iii), it vanishes along

γ[I]. And therefore, by (ii), it vanishes on spacelike hypersurfaces that intersect γ[I]. So it vanishes 1

2

3

everywhere. Thus, as claimed, we can find three smooth, spacelike fields σ a , σ a , σ a on some open set O containing p that satisfy the three listed conditions. And the fields must certainly be linearly independent throughout O — because we started with three linearly independent vectors at p, and linear independence is preserved under parallel transport. i

It follows from (ii), of course, that ∇[a σ b] = 0 for all i. So, restricting O to a smaller open set i

i

i

i

containing p if necessary, we can find smooth scalar fields x on O such that σa = ∇a x = hab ∇b x. (Here

we invoke proposition 4.1.6.) We can pin them down uniquely by requiring that they assume the value 0 at points on γ[I]. This guarantees that i

ηbn ∇n x = 0

on γ[I] ∩ O for all i. And, by condition (i),

(4.5.1)

j

i

hab (∇a x) (∇b x) = δij

(4.5.2) j

holds everywhere for all i and j. (Why? Contracting (i) with (∇b x) yields j

j

σ a = hab (∇b x) =

3 X i=1

1

2

j

σ a (σ b ∇b x). i

i

3

j

i

But the vectors σ a , σ a , σ a are linearly independent at every point. So σ b ∇b x = δij and, therefore, i

j

i

j

hab (∇a x)(∇b x) = σ b ∇b x = δij .)

Now we extend the tangent field ηba to a smooth field η a on O by requiring that ta η a = 1 and 3

2

1

i

η n ∇n x = 0 hold everywhere for all i. (The fields ta , (∇a x), (∇a x), (∇a x) form a co-basis at every point,

and so a vector field is uniquely determined by its contractions with them.) We claim that the resultant i

i

field η a is rigid, i.e., £η hab = 0. We have £η x = η n ∇n x = 0 for all i. And £η (∇a ϕ) = ∇a (£η ϕ) for all i

i

smooth scalar fields ϕ. (This is easily checked using proposition 1.7.4.) So £η (∇a x) = ∇a (£η x) = 0 for all i. Hence, by (4.5.2),

i

j

j

i

0 = £η (hab (∇a x) (∇b x)) = (∇a x) (∇b x) £η hab for all i and j. But we also have £η ta = η n ∇n ta +tn ∇a η n = 0 and, therefore, ta £η hab = £η (ta hab ) = 0. 1

2

3

Thus, contracting £η hab with any of the basis elements ta , (∇a x), (∇a x), (∇a x) yields 0. So £η hab = 0,

as claimed. Now we turn to the recovery theorem.


258

Proposition 4.5.2. (K¨ unzle-Ehlers Recovery Theorem) Let (M, ta , hab , ∇) be a classical spacetime that satisfies

Rbc Ra b c d

=

4 π ρ tbc ,

(4.5.3)

=

Rc d a b ,

(4.5.4)

for some smooth scalar field ρ on M . Let η a be a smooth, future-directed, unit timelike vector field on some open subset O of M that is rigid. (Existence of such fields, at least locally, is guaranteed by the ˆ ab be the projection field associated with η a , and let φa preceding proposition and proposition 4.1.5.) Let h and ωab be the associated acceleration and rotation fields: φa

= η n ∇n η a , ˆ m [a h ˆ b]n ∇m η n . = h

ωab

f

Then there exists a unique derivative operator ∇ on O such that all the following hold on O. f

(RR1) ∇ is compatible with ta and hab . f

f

(RR2) η a constant with respect to ∇ (i.e., ∇a η b = 0). f

(RR3) ∇ is flat. (RR4) For all timelike curves with four-velocity field ξ a , f

ξ n ∇n ξ a = 0 ⇐⇒ ξ n ∇n ξ a = −φa − 2 ωn a ξ n . (RR5) φa and ωab satisfy the “field equations”: f

[a bc] ∇ ω f

∇a ω

ab

f

=

0,

=

0,

[a b] ∇ φ

=

∇a φ

=

f

a

(4.5.5) (4.5.6) f

η n ∇n ω ab ,

4πρ − ωab ω ab .

(4.5.7) (4.5.8)

Note that, as promised, the Trautman Recovery Theorem emerges as a corollary. If we add the supplemental condition (Rab cd = 0), then, by proposition 4.2.4 again, we can find timelike fields locally that are rigid and twist-free. But if ωab = 0, it follows from (4.5.7) (and proposition 4.1.6) that φa must, at least locally, be of the form φa = ∇a φ for some smooth scalar field φ. And in this case (ωab = 0 and φa = ∇a φ), we fully recover the conclusions of the Trautman theorem.

The de-geometrization presented here is relativized to a rigid unit timelike vector field η a . Given that field, there is a unique derivative operator satisfying the listed conditions (relative to it). But it will be clear from the proof that, in general, different choices for η a lead to different derivative operators, i.e., lead to different de-geometrizations. Indeed, one has, here, much the same non-uniqueness that we encountered in the Trautman Recovery Theorem.

259


Proof. The argument here is similar in structure to the one we gave for the Trautman Recovery Theorem, and many individual steps carry over intact or with only minimal change. We just have to remember that whereas previously we had the condition ∇a η b = 0 to work with, we now have only ∇(a η b) = 0. Consider the fields κab

=

ˆ n[b ∇a] η n , h

C a bc

=

2 t(b κc) a

(4.5.9) (4.5.10)

on O. It is easy to check that they satisfy the following conditions. 2 κa b

=

ˆ na ∇b η n , ∇a η b − h

(4.5.11)

κab

=

∇a η b = ω ab ,

(4.5.12)

2 κa b η a

=

η a ∇a η b = φb ,

(4.5.13)

2 κa b

=

2 ∇a η b − ta φb ,

(4.5.14)

C a ac

=

0.

(4.5.15)

We get the second from the fact that η a is rigid, and so ∇a η b = ∇[a η b] = ω ab . The fourth follows from

the second and third. (Note that contracting both sides with either har or η a yields the same result.) The fifth follows from the anti-symmetry of κab . f

Next consider the derivative operator ∇ = (∇, C a bc ) on O. We claim that it satisfies all the listed

conditions. (RR1) follows immediately from proposition 4.1.3. For (RR2), note that, by (4.5.13) and (4.5.14), f

b ∇a η

= ∇a η b − C b an η n = ∇a η b − (ta κn b + tn κa b ) η n 1 1 = ∇a η b − ta φb − ∇a η b − ta φb = 0. 2 2 f

Thus, as required for (RR2), η a is constant with respect to the new derivative operator ∇. f

Now we turn to the Riemann curvature field associated with ∇. We have, by (1.8.2), f

R

a

bcd

= Ra bcd + 2 ∇[c C a d]b + 2 C n b[c C a d]n = Ra bcd + 2 t[d ∇c] κb a + 2 tb ∇[c κd] a + 2 tb t[d κc] n κn a .

(4.5.16)

f

It follows immediately that Rabcd = Rabcd . But Rabcd = 0. (By proposition 4.1.5, this is a consequence f

of the geometrized version of Poisson’s equation (4.5.3).) So ∇ is spatially flat. Now recall the second

clause of proposition 4.2.4. We have just verified that there is a smooth, unit timelike field η a on O that f

f

f

is constant with respect to ∇. So (since ∇ is spatially flat), the proposition tells us that ∇ must be flat f

outright, i.e., Ra bcd = 0. So we have (RR3). And (4.5.16) reduces to

Ra bcd = −2 t[d ∇c] κb a − 2 tb ∇[c κd] a − 2 tb t[d κc] n κn a .

(4.5.17)

260


For (RR4), note first that for all timelike curves on O with four-velocity field ξ a , f

ξ n ∇n ξ a

=

ξ n (∇n ξ a − C a nm ξ m ) = ξ n ∇n ξ a − (tm κn a + tn κm a )ξ m ξ n

=

ξ n ∇n ξ a − 2 κn a ξ n .

But, by (4.5.14), (4.1.42), and the fact that η a is rigid (θn a = 0), 2 κn a

=

2 ∇n η a − tn φa

= 2 (ωn a + tn φa ) − tn φa = 2 ωn a + tn φa .

(4.5.18)

f

So ξ n ∇n ξ a = 0 iff ξ n ∇n ξ a = −φa − 2 ωn a ξ n . Thus we have (RR4). f

Notice also that there can be at most one derivative operator ∇ on O satisfying condition (RR4), so f

f

we get our uniqueness claim. For suppose that ∇′ = (∇, C ′a bc ) satisfies it as well. Then, for all timelike

geodesics on O (with respect to ∇) with four-velocity field ξ a , we have f

f

f

ξ n ∇n ξ a = −φa − 2 ωn a ξ n = ξ n ∇ ′ n ξ a = ξ n (∇n ξ a − C ′a nm ξ m ). So, C ′a nm ξ m ξ n = 0 holds at every point. But every future-directed unit timelike vector ξ a at a point in O is the tangent vector of some geodesic (with respect to ∇) through the point, and the collection

of future-directed unit timelike vectors at a point spans the tangent space there. So it follows that C ′a nm = 0 at every point in O.

Now, finally, we turn to (RR5). The four conditions we must verify all follow from (4.5.17). Contracting a with d yields 4 π ρ tb tc = Rbc = tc ∇a κb a + tb ∇a κc a + tb tc κa n κn a .

(4.5.19)

Ra b c d = −td ∇c κb a − tb ∇c κd a + tb ∇d κca − tb td κcn κn a .

(4.5.20)

And raising ‘c’ yields

Let us now contract (4.5.19) with η b hcr . This, together with (4.5.12), gives us 0 = ∇a κra = ∇a ω ra . It follows that

f

ab ab nb a an b ∇a ω = ∇a ω − ω C an − ω C an = 0.

(4.5.21)

(Here C a an = 0 by (4.5.15), and ω an C b an = 0 because of the respective anti-symmetry and symmetry of ω an and C b an .) So we have the second in our list of four (RR5) conditions. Next, let us contract (4.5.19) with η b η c . Then, using (4.5.13), (4.5.14), and (4.5.18), we get 4πρ

= 2 η b ∇a κb a + κa n κn a = 2 [∇a (κb a η b ) − (∇a η b )κb a ] + κa n κn a 1 = ∇a φa − 2 (κa b + ta φb )κb a + κa n κn a = ∇a φa − κa b κb a = ∇a φa − ωa b ωb a . 2

So, by (4.5.15) again, f

a a a n a ab ∇a φ = ∇a φ − C an φ = ∇a φ = 4 π ρ − ωab ω .

(4.5.22)

261


Thus we have the fourth condition in the (RR5) list. That leaves the first and the third. Now, for the first time, we use the fact that ∇ satisfies the first supplemental curvature condition f

f

(4.5.4). Since ∇ satisfies it as well — as it clearly does since R a bcd = 0 — we know from proposition f

f

f

4.3.5 that κab must be closed, i.e, ∇ [a κbc] = 0. So, by (4.5.12), ∇ [a ω bc] =∇ [a κbc] = 0. That is the first

condition in the list. Finally, contracting (4.5.20) with η b η d and using (4.5.13), (4.5.12), and (4.5.18) yields Ra b c d η b η d

= −2 η b ∇c κb a + η d ∇d κca − κcn κn a = −2 ∇c (κb a η b ) − (∇c η b )κb a + η d ∇d κca − κcn κn a

=

−∇c φa + η d ∇d κca + κcb κb a = −∇c φa + η d ∇d ω ca + ω cb ωb a .

So, since Ra b c d = Rc d a b (and since ω cn is anti-symmetric), ∇[a φc] − η d ∇d ω ac = 0. But, as one can easily check (with a computation much like ones we have seen before), f

f

∇ a φc = ∇a φc

and

η d ∇ d ω ac = η d ∇d ω ac . This gives us the third condition in the (RR5) list, and we are done.

We have claimed that the difference between the two versions of geometrized Newtonian gravitation theory collapses if one restricts attention to classical spacetimes that are, in a certain weak sense, “asymptotically flat”. (In that case, the second supplemental curvature condition, Rab cd = 0, follows from the other assumptions.) Now we make the claim precise. Toward that goal, we first prove a result of Ehlers’ [15]. Proposition 4.5.3. Let (M, ta , hab , ∇) be a classical spacetime that is spatially flat (Rabcd = 0). Then there is a smooth scalar field Ψ on M such that

Moreover,

Ra bcd Rb a c e = Ψ td te .

(4.5.23)

Rab cd = 0 ⇐⇒ Ψ = 0.

(4.5.24)

Proof. Let p be any point in M , let η a be a smooth, future-directed, rigid, unit timelike field defined on some open set containing p — existence is guaranteed, once again, by proposition 4.5.1 — and let ωab be f

the rotation field determined by η a . Further, let κab and ∇ be defined (relative to η a ) as in the preceding

proof. Then, by (4.5.17), (4.5.20), and (4.5.18), Ra bcd Rb a c e

=

−2 t[d ∇c] κb a − 2 tb ∇[c κd] a − 2 tb t[d κc] n κn a

−te ∇c κa b − ta ∇c κe b + ta ∇e κcb − ta te κcn κn b = (−td ∇c κb a )(−te ∇c κa b ) = td te (∇c ωb a )(∇c ωa b ).

So we need only take Ψ = −(∇c ωab )(∇c ω ab ) at p. Now, by (4.5.17) again, we also have Rab cd =

−2 t[d ∇c] κba = −2 t[d ∇c] ω ba . Hence (since contracting Rab cd with either η c η d or hcr hds yields 0), Rab cd = 0 ⇐⇒ ∇c ω ba = 0.

(4.5.25)

262


So the assertion that remains for us to prove, namely (4.5.24), is equivalent to ∇c ω ba = 0 ⇐⇒ (∇c ωab )(∇c ω ab ) = 0.

(4.5.26)

One direction is trivial, of course. And the other (right to left) follows just from the fact that the indices in ∇c ω ba are spacelike (and the metric induced by hab on the space of spacelike vectors at any point is 1

2

3

positive definite). For future reference, we give the argument in detail. Let σ a , σ a , σ a be three linearly

independent, smooth spacelike fields on some open set containing p such that (i) hab = i

P3

i=1

i

i

σa σ b

(ii) ∇a σ b = 0. (Existence is guaranteed by our assumption of spatial flatness. Recall the proof of proposition 4.5.1.) Let 1

2

3

i

i

i

j

λa , λa , λa be three smooth fields such that σ a = hab λb (or, equivalently, σ a λa = δij ) for all i and j. Now, ijk

for all i, j, and k, let ω be the scalar field defined by ijk

j

i

k

j

ω = σ c σ a σ b (∇c ωab ) = λc λa λb (∇c ω ab ). k

i

Then ∇c ω ab =

3 X

ijk

j

ω σc σa σ b , k

i

i,j,k=1

and, hence, (∇c ωab )(∇c ω ab ) =

3 X

ijk

i,j,k=1

j

ω (σ c σ a σ b ∇c ωab ) = k

i

3 X

ijk

( ω )2 .

(4.5.27)

i,j,k=1

ijk

So, clearly, (∇c ωab )(∇c ω ab ) can vanish only if ω = 0 for all i, j, k, i.e., only if ∇c ω ab = 0. Now we can formulate our notion of asymptotic flatness. It is intended to capture the intuitive claim that “Rab cd goes to 0 at spatial infinity”. (We could certainly impose a restriction on the limiting behavior of Ra bcd but, in fact, it suffices for our purposes to work with a weaker condition that is formulated in terms of Rab cd .) With equivalence (4.5.24) in mind, we shall use the condition Ψ → 0 as a surrogate for

the condition Rab cd → 0.

We first have to insure that there is an asymptotic regime in which spacetime curvature can go (or fail to go) to zero. We do so by restricting attention to classical spacetimes that can be foliated by a family of spacelike hypersurfaces that are simply connected and geodesically complete. Each of these hypersurfaces (together with the metric induced on it by hab ) is then, in effect, a copy of ordinary three-dimensional Euclidean space. Given a classical spacetime (M, ta , hab , ∇) satisfying this condition, we say officially

that Rab cd goes to 0 at spatial infinity if, for all spacelike geodesics γ : R → M , Ψ(γ(s)) → 0 as s → ∞. Now we can formulate the collapse result (due to K¨ unzle [35] and Ehlers [15]).

Proposition 4.5.4. Let (M, ta , hab , ∇) be a classical spacetime that is spatially flat. Suppose the following

conditions hold.

263


(1) For all p in M , there is a spacelike hypersurface containing p that is simply connected and geodesically complete. (2) Rab cd goes to 0 at spatial infinity (in the sense discussed above). Then Rab cd = 0 (everywhere). Proof. Arguing as in the proof of proposition 4.5.1, but now using assumption (1), we can show that 1

2

3

there exist three smooth, linearly independent, globally defined spacelike fields σ a , σ a , σ a satisfying 1

1

2

2

3

3

(i) hab = σ a σ b + σ a σ b + σ a σ b i

(ii) ∇a σ b = 0, and there exists a smooth globally defined future-directed, unit timelike field ξ a that is rigid (∇(a ξ b) = 0). f

Let ωab be the rotation field associated with the latter, and let ∇ be its associated flat derivative operator ijk

(as constructed in the proof of proposition 4.5.2). Finally, let the scalar component fields ω be defined by ijk

j

ω = σ c σ a σ b (∇c ωab ) k

i

as in the preceding proof. We are assuming that Ψ = (∇c ωab )(∇c ω ab ) goes to 0 as one approaches P3 ijk spatial infinity. But, by (4.5.27), (∇c ωab )(∇c ω ab ) = i,j,k=1( ω )2 . Hence, for all i, j, and k, ijk

(a) ω → 0 at spatial infinity. ijk

We claim now that the fields ω are all harmonic, i.e., f

f

ijk

(b) ∇n ∇n ω = 0.

ijk

(We could equally well take the claim to be ∇n ∇n ω = 0, but it is more convenient to work with the flat f

derivative operator ∇.) Once we show this, we will be done. Because it will then follow by the “minimum P ijk ijk principle” that the fields ω all vanish.6 That, in turn, will imply that Ψ = 3i,j,k=1( ω )2 = 0 and, hence, by (4.5.24), that Rab cd = 0.

1

2

3

i

f

i

As in the preceding proof, let λa , λa , λa be three smooth fields such that σ a = hab λb . Now ∇a and ∇a

agree in their action on contravariant fields that are spacelike in all indices. In particular, for all i, f

∇a σ b

=

∇a σ b ,

∇a ω bc

=

∇a ω bc .

i

f

6 See,

i

e.g., Flanders [20], p. 85. The principle asserts that a harmonic function defined on a compact set in three-

dimensional Euclidean space assumes its minimum value on its boundary. It follows — consider a nested sequence of closed balls with radii going to infinity — that if a harmonic function defined on all of three-dimensional Euclidean space goes to 0 asymptotically along any (or even just one) geodesic, then it must be 0 everywhere. ijk

We here apply the principle to the fields ω or, rather, the restrictions of those fields to individual spacelike hypersurfaces that are simply connected and geodesically complete. Note that condition (b) can be construed as a constraint on the f

restricted fields. If D is the (three-dimensional) derivative operator induced on a spaceike hypersurface by ∇ — which is ijk

the same as the one induced by ∇ — then it follows from (b) that Dn D n ω = 0.

264


f

i

i

i

(Here C a bc has the form C a bc = 2 t(b κc) a . So C ab c = tc κba and, hence, ∇a σ b = ∇a σ b − C ba m σ m = i

i

f

i

ijk

i

∇a σ b − tm κab σ m = ∇a σ b . The other case is handled similarly.) It follows that ∇a σ b = 0, and ω = k

i

f

j

k

i

j

f

f

f

λc λa λb (∇c ω ab ) = σ c σ a σ b (∇c ωab ). Hence, f

f

ijk

∇n ∇n ω

f

j

f

f

f

j

f

f

j

f

= ∇n ∇n (σ c σ a σ b ∇c ωab ) = ∇n (σ c σ a σ b ∇n ∇c ωab ) = ∇n (σ c σ a σ b ∇n ∇c ωab ) k

j

f

i

f

f

k

k

f

j

i

i

f

k

i

f

= σ c σ a σ b (∇n ∇n ∇c ωab ) = λc λa λb (∇n ∇n ∇c ω ab ) k

i

for all i, j, and k. So, to complete the proof, it suffices for us to show f

f

f

(c) ∇n ∇n ∇c ω ab = 0.

f

f

And this condition follows easily from the fact that ∇ [a ω bc] = 0 and ∇a ω ab = 0. (Recall (4.5.5) and f

(4.5.6) in the formulation of proposition 4.5.2.) Since ∇ is flat, we can switch derivative operator position and, therefore, f

f

f

f

f

f

f

f

f

f

f

f

f

f

f

f

∇n ∇n ∇c ω ab = ∇c ∇n ∇n ω ab = ∇c ∇n (−∇b ω na − ∇a ω bn ) = −∇c ∇b (∇n ω na ) − ∇c ∇a (∇n ω bn ) = 0. So we are done.

Solutions to Problems b be an open subset Problem 1.1.1 Let (M, C) be an n−manifold, let (U, ϕ) be an n−chart in C, let O b Show that (O, ϕ|O ) is also an n−chart in C. of ϕ[U ], and let O be its preimage ϕ−1 [O].

Let ϕ′ be the restricted map ϕ|O . (We write it this way just to simplify our notation.) Clearly, ϕ′ [O] b And ϕ′ is one-to-one (since it is a restriction of ϕ). So (O, ϕ′ ) qualifies is open, since ϕ′ [O] = ϕ[O] = O.

as an n-chart on M . To show that it belongs to C, we must verify that it is compatible with every n-chart in C.

Let (V, ψ) be one such. We may assume that U ∩ V is non-empty, since otherwise the charts are

automatically compatible. Since ϕ′ is a restriction of ϕ, and O is a subset of U (and ϕ is one-to-one), we have ϕ′ [O ∩ V ] = ϕ[O ∩ V ] = ϕ[O ∩ (U ∩ V )] = ϕ[O] ∩ ϕ[U ∩ V ].

b and ϕ[U ∩ V ] is open (since the charts (U, ϕ) and (V, ψ) are But ϕ[O] is open (since it is equal to O),

compatible). So ϕ′ [O ∩ V ] is open. Furthermore, ψ[O ∩ V ] is open since it is the preimage of the open set ϕ[O ∩ V ] under the smooth (hence continuous) map

ϕ ◦ ψ −1 : ψ[U ∩ V ] → ϕ[U ∩ V ]. (That the map is smooth follows, again, by the compatibility of the charts (U, ϕ) and (V, ψ).) Finally, the maps ϕ′ ◦ ψ −1 : ψ[O ∩ V ] → ϕ′ [O ∩ V ], ψ ◦ ϕ′−1 : ϕ′ [O ∩ V ] → ψ[O ∩ V ] are smooth since they are the restrictions to open sets, respectively, of the smooth maps ϕ ◦ ψ −1 : ψ[U ∩ V ] → ϕ[U ∩ V ], ψ ◦ ϕ−1 : ϕ[U ∩ V ] → ψ[U ∩ V ]. Problem 1.1.2 Let (M, C) be an n−manifold, let (U, ϕ) be an n−chart in C, and let O be an open set in M such that U ∩ O 6= ∅. Show that U ∩ O, ϕ|U∩O is also an n−chart in C. 265

SOLUTIONS TO PROBLEMS

266

We claim, first, that ϕ[U ∩ O] is open. To see this, let ϕ(p) be any point in ϕ[U ∩ O]. Since O is open,

there exists an n-chart (V, ψ) in C where p ∈ V ⊆ O. Since (V, ψ) and (U, ϕ) are compatible, ϕ[U ∩ V ]

qualifies as an open subset of ϕ[U ∩ O] containing ϕ(p). So ϕ[U ∩ O] is open, as claimed. It now follows b = ϕ[U ∩ O]) that the pair U ∩ O, ϕ|U∩O is an n-chart in C. by the result of problem 1.1.1 (taking O Problem 1.1.3 Let (M, C) be an n−manifold and let T be the set of open subsets of M . (i) Show that T

is a topology on M , i.e., it contains the empty set and the set M , and is closed under finite intersections and arbitrary unions. (ii) Show that T is the coarsest topology on M with respect to which ϕ : U → Rn

is continuous for all n−charts (U, ϕ) in C.

(i) The empty set qualifies, vacuously, as open, and M qualifies as open since (M, C) satisfies condition

(M2). So we need only show that T is closed under finite intersections and arbitrary unions. For the first claim, it suffices to show that if O1 and O2 are both open, then their intersection O1 ∩ O2 is as well. (The

claim will then follow by induction.) So assume that O1 and O2 are open, and let p be a point in O1 ∩ O2 .

(If the intersection is empty, it is automatically open.) Since O2 is open, there is an n-chart (U, ϕ) in C such that p ∈ U ⊆ O2 . Then, by the result in problem 1.1.2, the pair U ∩ O1 , ϕ|U∩O1 is an n-chart in C. Thus, given an arbitrary point p in O1 ∩ O2 , there is an n-chart in C (namely, (U ∩ O1 , ϕ|U∩O1 ))

whose domain contains p and is a subset of O1 ∩ O2 . It follows that O1 ∩ O2 is open, as claimed. Finally,

let S be a set of open sets, and let p be a point in its union ∪S. (Again, if the union is empty, it is automatically open.) Let O be a set in S such that p ∈ O. Since O is open, there is an n-chart (U, φ)

in C such that p ∈ U ⊆ O ⊆ (∪S). So, given our arbitrary point in ∪S, there is an n-chart in C (namely

(U, ϕ)) whose domain contains p and is a subset of ∪S. It follows that ∪S is open.

(ii) First, we claim that given any n-chart (U, ϕ) in C, ϕ : U → Rn is continuous with respect to T . b of ϕ[U ], its preimage ϕ−1 [O] b is Let (U, ϕ) be one such. We need to show that, given any open subset O

b open. But by the result in problem 1.1.1, we know that there is an n-chart in C whose domain is ϕ−1 [O].

And the domain of an n-chart in C is certainly open. So our claim follows easily. Next, assume that T ′

is a topology on M with respect to which ϕ : U → Rn is continuous for all n−charts (U, ϕ) in C. We

show that T ⊆ T ′ . Let O be a set in T , and let p be a point in O. (If O is empty, then it certainly

belongs to T ′ since the latter is a topology on M .) Since O is open, there is an n-chart (U, ϕ) in C such

that p ∈ U ⊆ O. By assumption, ϕ is continuous with respect to T ′ . And ϕ[U ] is an open set in Rn (by

the defintion of an n-chart). So its preimage U must belong to T ′ . Thus given any point p in O, there

is a T ′ -open set (namely, U ) that contains p and is a subset of O. It follows that O itself is open with

respect to T ′ . Thus, as claimed, every set O that belongs to T belongs to T ′ as well.

Problem 1.1.4 Let (M, C) be an n−manifold. Show that a map α : M → R is smooth according to our first definition of “smoothness” (which applies only to real-valued maps on manifolds) iff it is smooth according to our second definition (which applies to maps between arbitrary manifolds). α is smooth in the first sense iff for all n-charts (U, ϕ) in C, the map α ◦ ϕ−1 : ϕ[U ] → R is smooth. It

267


is smooth in the second sense iff for all smooth maps β : R → R, the composed map β ◦ α : M → R is smooth in the first sense (i.e., (β ◦ α) ◦ ϕ−1 : ϕ[U ] → R is smooth for all n-charts (U, ϕ) in C). To see that

the second sense implies the first, we need only consider the special case where β is the identity map on R.

For the converse, suppose that α is smooth in the first sense, let β : R → R be any smooth map on R, and let (U, ϕ) be any n-chart in C. Then (β ◦ α) ◦ ϕ−1 : ϕ[U ] → R is smooth since (β ◦ α) ◦ ϕ−1 = β ◦ (α ◦ ϕ−1 ),

i.e., it is composition of smooth maps α ◦ ϕ−1 : ϕ[U ] → R and β : R → R.

In what follows, let (M, C) be an n−manifold, let p be a point in M , and let C(p) be the set of charts

in C whose domains contain p.

Problem 1.2.1 Let ξ be a non-zero vector at p, and let (k 1 , ..., k n ) be a non-zero element of Rn . Show there exists an n-chart in C(p) with respect to which ξ has components (k 1 , ..., k n ). Let (U1 , ϕ1 ) be an n-chart in C(p), and let (ξ 1 , ..., ξ n ) be the components of ξ with respect to (U1 , ϕ1 ).

These components cannot all be 0, since ξ is not the zero vector. So there is an isomorphism L of (the

vector space) Rn onto itself that takes (ξ 1 , ..., ξ n ) to (k 1 , ..., k n ). Let its associated matrix have elements n X aij ξ j . {aij }. Then, for all i = 1, ..., n, k i = j=1

Now consider a new n-chart (U2 , ϕ2 ) in C(p) where U2 = U1 and ϕ2 = L ◦ ϕ1: U2 → Rn . (That it is an

n-chart and does belong to C must be checked. But these claims follow easily from the fact that L, now construed as a map from the manifold Rn to itself, is a diffeomorphism.) We claim that the components

of ξ with respect to (U2 , ϕ2 ) are (k 1 , ..., k n ). To see this, we invoke proposition 1.2.5. For all i = 1, ..., n, let x′i : ϕ1 [U1 ∩ U2 ] → R be the coordinate map defined by x′i = xi ◦ ϕ2 ◦ ϕ−1 1 . Since ϕ2 = L ◦ ϕ1 , we

have

xi ◦ ϕ2 = and, therefore,

n X j=1

aij (xj ◦ ϕ1 )

x′i = xi ◦ ϕ2 ◦ ϕ1−1 =

n X

aij xj .

j=1

It now follows by proposition 1.2.5 that the components of ξ with respect to (U2 , ϕ2 ) are ξ ′i =

n X j=1

n

ξj

X ∂x′i ξ j aij = k i , (ϕ (p)) = 1 ∂xj j=1

for all i. ∂ ∂ − x2 2 on R2 . Show that the maximal integral curve ∂x1 ∂x of ξ with initial value p = (p1 , p2 ) is the map γ : R → R2 with γ(s) = (p1 es , p2 e−s ). Problem 1.3.1 Let ξ be the vector field x1

γ has initial value (p1 , p2 ). It is an integral curve of the given vector field since, for all s ∈ R, and all

268


f ∈ S(γ(s)), by the chain rule, ~γγ(s) (f ) = = = =

d d f ◦ γ (s) = f p1 es , p2 e−s ) ds ds

∂f ∂f (γ(s)) (p1 es ) + 2 (γ(s)) (−p2 e−s ) 1 ∂x ∂x ∂f ∂f (γ(s)) x1 (γ(s)) − 2 (γ(s)) x2 (γ(s)) ∂x1 ∂x 1 ∂ 2 ∂ x −x (f ). ∂x1 ∂x2 |γ(s)

Finally, it is maximal because its domain is R. Problem 1.3.2 Let ξ be a smooth vector field on M , let p be a point in M , and let s0 be any real number (not necessarily 0). Show that there is an integral curve γ : I → M of ξ with γ(s0 ) = p that is

maximal in the sense that given any integral curve γ ′ : I ′ → M of ξ, if γ ′ (s0 ) = p, then I ′ ⊆ I and

γ ′ (s) = γ(s) for all s in I ′ .

Given an interval J, let us understand J +a to be the translation of J by the number a. Let σ : J → M

be the maximal integral curve of ξ with initial value p. (Existence is guaranteed by proposition 1.3.1.) Let I be the shifted interval J + s0 , and let γ : I → M be the curve defined by γ(s) = σ(s − s0 ). Then

γ is an integral curve of ξ by the first clause of proposition 1.3.2, and γ(s0 ) = σ(0) = p. We claim that γ satisfies the stated maximality condition.

To see this, suppose γ ′ : I ′ → M is an integral curve of ξ, and γ ′ (s0 ) = p. Let J ′ = I ′ − s0 and

let σ ′ : J ′ → M be defined by σ ′ (s) = γ ′ (s + s0 ). Then σ ′ is an integral curve of ξ (by the first clause

of proposition 1.3.2 again) with initial value 0 (since σ ′ (0) = γ ′ (s0 ) = p). So, by the maximality of σ,

J ′ ⊆ J and σ ′ (s) = σ(s) for all s in J ′ . It follows immediately that I ′ = J ′ + s0 ⊆ J + s0 = I and γ ′ (s) = σ ′ (s − s0 ) = σ(s − s0 ) = γ(s) for all s in I ′ .

Problem 1.3.3 (Integral curves that go nowhere) Let ξ be a smooth vector field on M , and let γ : I → M

be an integral curve of ξ. Suppose that ξ vanishes (i.e., assigns the zero vector) at some point p ∈ γ[I].

Then the following both hold.

(1) γ(s) = p for all s in I (i.e., γ is a constant curve). (2) The reparametrized curve γ ′ = γ ◦ α : I ′ → M is an integral curve of ξ for all diffeomorphisms α : I ′ → I.

(1) Suppose s0 ∈ I and γ(s0 ) = p. It follows from problem 1.3.2 that there is a unique maximal

integral curve of ξ whose value at s0 is p. The only possibility is the constant curve γˆ : R → M that assigns p to all s. (ˆ γ is an integral curve of ξ since, for all f ∈ S(p), f ◦ γˆ is constant and, so, ~γˆ|ˆγ (s) (f ) = d (f ◦ γˆ)(s) = 0 = ξ|ˆγ (s) (f ) ds

269


for all s. It is maximal since its domain is R.) Hence, by maximality, γ(s) = γˆ(s) = p for all s in I. (2) Let α : I ′ → I be a diffeomorphism and let γ ′ be the composed map γ ′ = γ ◦ α : I ′ → M . We know

from (1.3.3) that γ ′ is an integral curve of ξ iff

dα ξ γ(α(s)) (s) = ξ γ(α(s)) ds

for all s in I ′ . But γ(α(s)) = p for all s in I ′ (by the first part of the problem) and, therefore, ξ γ(α(s)) = ξ(p) = 0 for all s in I ′ . So the required equation holds for all s in I ′ . (Both sides are 0.)

Problem 1.3.4 (Integral curves cannot cross) Let γ : I → M and γ ′ : I ′ → M be integral curves of ξ

that are maximal (in the sense of problem 1.3.2) and satisfy γ(s0 ) = γ ′ (s′0 ). Then the two curves agree up to a parameter shift: γ(s) = γ ′ s + (s′0 − s0 ) for all s in I. Let I ′′ = I ′ − (s′0 − s0 ), and let γ ′′ : I ′′ → M be the curve defined by γ ′′ (s) = γ ′ s + (s′0 − s0 ) .

It is an integral curve of ξ by proposition 1.3.2, and γ ′′ (s0 ) = γ ′ (s′0 ) = γ(s0 ). So by the maximality of γ, I ′′ ⊆ I and γ ′′ (s) = γ(s) for all s in I ′′ , i.e., γ(s) = γ ′ (s + (s′0 − s0 )) for all s in I ′′ . It remains to

verify only that I ′′ = I. Since I ′′ ⊆ I, it follows that I ′ ⊆ I + (s′0 − s0 ). If we rerun the argument with

the roles of I, γ, and s0 interchanged with those of I ′ , γ ′ , and s′0 , we arrive at the symmetric conclusion that I ⊆ I ′ + (s0 − s′0 ). Putting the two set inclusions together, we arrive at I ⊆ I ′ + (s0 − s′0 ) ⊆

I + (s′0 − s0 ) + (s0 − s′0 ) = I. So I ′′ = I ′ + (s0 − s′0 ) = I, as claimed.

Problem 1.3.5 Let ξ be a smooth vector field on M that is complete. Let p be a point in M . Show that the restriction of ξ to the punctured set M − {p} is complete (as a field on M − {p}) iff ξ vanishes at p. Let ξ ′ be the restriction of ξ to M − {p}. Suppose first that ξ vanishes at p. Then, as we know from

problem 1.3.3, every integral curve of ξ that passes through p is necessarily a degenerate constant curve that sits at p. It follows, we claim, that ξ ′ is complete. For let q be any point in M distinct from p. Since ξ is complete (as a field on M ), there is an integral curve γ : R → M of ξ with initial value q. The image

of γ is fully contained in M − {p} (since otherwise γ would be an integral curve of ξ passing through p that does not sit at p). So γ qualifies as an integral curve of ξ ′ . Since the domain of γ is R (and since q was chosen arbitrarily), we see that ξ ′ is complete, as claimed.

Conversely, suppose ξ does not vanish at p. Since ξ is complete (as a field on M ), there is an integral curve γ : R → M of ξ with initial value p. γ cannot be a constant curve that sits at p. (Otherwise, we

would have ~γp = 0 and, hence, ξ(p) = 0.) So the set D = {s ∈ R : γ(s) 6= p} is non-empty. It is a disjoint

union of open intervals. (If 0 is the only number s in R such that γ(s) = p, then D will be the union of (−∞, 0) and (0, ∞). Other possibilities arise because γ may pass through p more than once.) Let I ′ be

any one of these intervals, let γ ′ : I ′ → M be the restriction of γ to I ′ , and let q be any point in γ ′ [I ′ ].

Then γ ′ qualifies as a maximal integral curve of ξ ′ in M − {p} that passes through q . By shifting initial

270


values, we can generate a maximal integral curve γ ′′ of ξ ′ in M − {p} that has initial value q. But the

domain of γ ′′ is not R (since the pre-shifted domain I ′ of γ ′ is not R). So we may conclude that ξ ′ is not complete. Problem 1.4.1 Show that lemma 1.4.1 can also be derived as a corollary to the following fact about square matrices: if M is an (r × r) matrix (r ≥ 1) and M × M is the zero matrix, then the trace of M

is 0.

Assume the left-side condition

r P

k=1 i

j

k

ϕ a ψ c = 0 holds, and let M be the r × r matrix with entries k

Mij = ϕa ψ a . Then M × M is the zero matrix since (M × M )ij =

r X

Mik Mkj =

k=1

r X

k

i

k

r P

Mkk =

k=1

i

j

r X

k

(ϕb ψ a ) = 0. k

k=1

k=1

So, by the stated fact, 0 = tr(M ) =

j

(ϕa ψ a )(ϕb ψ b ) = (ϕa ψ b ) r P

k=1

k

k

ϕa ψ a .

Problem 1.6.1 Show that for all smooth vector fields ξ a on M , £ξ δab = 0. For all smooth vector fields λa on M , we have λa £ξ δab = £ξ (δab λa ) − δab £ξ λa = £ξ λb − £ξ λb = 0. (The first equality follows from the Leibniz rule, and the second from the fact that δab functions as an index substitution operator.) Since this holds for all smooth fields λa (at all points in M ), we may conclude that £ξ δab = 0. Here is a second argument. By the Leibniz rule, and the fact that δab functions as an index substitution operator, we have £ξ δab = £ξ (δcb δac ) = δcb £ξ δac + δac £ξ δcb = £ξ δab + £ξ δab . It follows immediately that £ξ δab = 0. Problem 1.6.2 Let ξ a and η a be smooth vector fields on M , and let the latter be non-vanishing. Show that if £ξ (η a η b ) = 0, then £ξ η a = 0. Assume that £ξ (η a η b ) = 0, and let p be any point in M . Since η a is non-vanishing, there exists a smooth field λa on M such that the scalar field η a λa is non-zero at p. At all points we have 0 = λa λb £ξ (η a η b ) = λa λb (η a £ξ η b + η b £ξ η a ) = 2(λa η a )λb £ξ η b . Hence, λb £ξ η b = 0 at p. But we also have 0 = λb £ξ (η a η b ) = λb (η a £ξ η b + η b £ξ η a ) = η a λb £ξ η b + (λb η b )£ξ η a at all points. So (λb η b )£ξ η a = 0 at p and, therefore, £ξ η a = 0 at p. Since p is an arbitrary point in M , we are done.

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Problem 1.6.3 Show that the set of smooth contravariant vector fields on M forms a Lie algebra under the bracket operation, i.e., show that for all smooth vector fields ξ, η, λ on M , [ξ, η] = −[η, ξ] and

λ, [ξ, η] + η, [λ, ξ] + ξ, [η, λ] = 0.

The anti-symmetry of the bracket operation is immediate. We can establish the second condition with a straightforward computation. Let ξ, η, λ be smooth contravariant vector fields on M , and let α be a smooth scalar field on M . Then

λ, [ξ, η] (α)

=

λ ([ξ, η](α)) − [ξ, η](λ(α))

=

[λ(ξ(η(α))) − λ(η(ξ(α)))] − [ξ(η(λ(α))) − η(ξ(λ(α)))] .

=

[η(λ(ξ(α))) − η(ξ(λ(α)))] − [λ(ξ(η(α))) − ξ(λ(η(α)))]

=

[ξ(η(λ(α))) − ξ(λ(η(α)))] − [η(λ(ξ(α))) − λ(η(ξ(α)))] .

Similarly,

η, [λ, ξ] (α) ξ, [η, λ] (α)

When we add the three lines, we get 0 on the right side because each term has a mate with the opposite sign. Since this holds for all smooth scalar fields α on M , we have our second claim. Problem 1.6.4 Show that for all smooth vector fields ξ a , η a on M , and all smooth scalar fields α on M,

£(α ξ) η a = α £ξ η a − £η α ξ a .

Given any smooth scalar field β on M , we have £(α ξ) η a (β)

= = = = =

(α ξ)(η(β)) − η ((α ξ)(β)) α ξ(η(β)) − α η(ξ(β)) − η(α) ξ(β) α ξ(η(β)) − η(ξ(β)) − η(α) ξ(β) α £ξ η a (β) − (£η α) ξ(β) α £ξ η a − (£η α) ξ a (β).

Since this is true for all smooth scalar fields β, it follows that £(α ξ) η a = α £ξ η a − £η α ξ a . Problem 1.6.5 One might be tempted to take a smooth tensor field to be “constant” if its Lie derivatives with respect to all smooth vector fields are zero. But this idea does not work. Any contravariant vector field constant in this sense would have to vanish everywhere. Prove this. Let η a be a smooth vector field on M . Assume that £ξ η a = 0 for all smooth vector fields ξ a on M . Then, given any smooth scalar field α on M , it follows from the preceding problem that 0 = £(α ξ) η a = α £ξ η a − £η α ξ a = − £η α ξ a .

272


Since this is true for all smooth vector fields ξ a on M , £η α = 0. Equivalently, η(α) = 0. But this is true for alll smooth scalar fields α on M . So η a = 0. Problem 1.6.6 Show that for all smooth vector fields ξ a , η a on M , and all smooth tensor fields αa...b c...d on M , where θa is the field £ξ η a .

...ar ...ar = £θ αba11...b £ξ £η − £η £ξ αba11...b s s

Consider first the case of a smooth scalar field α on M . The assertion follows since £ξ £η − £η £ξ (α) = ξ(η(α)) − η(ξ(α)) = (£ξ η)(α) = £θ α.

Next consider the case of a smooth vector field αa on M . Given any smooth scalar field β on M , we have

£ξ £η − £η £ξ αa (β)

=

£ξ £η αa (β) − £η £ξ αa (β)

=

[ξ, [η, α]](β) − [η, [ξ, α]](β)

=

−[α, [ξ, η]](β)

=

[[ξ, η], α](β) = (£θ αa )(β).

(Note that the third and fourth equalities follow from the assertions in problem 1.6.3.) Since this is true for all smooth scalar fields β, (£ξ £η − £η £ξ ) αa = £θ αa . The other cases now follow in standard

computational sequence. To compute (£ξ £η − £η £ξ ) αb , we consider an arbitrary smooth field λb and make use of our previous derived expressions for (£ξ £η − £η £ξ ) (αb λb ) and (£ξ £η − £η £ξ ) λb . And so forth.

Problem 1.7.1 Let ∇ be a derivative operator on a manifold. Show that ∇n δab = 0. We can use much the same argument here as used for Problem 1.6.1. By the Leibniz rule, and the fact that δab functions as an index substitution operator, ∇n δab = ∇n (δcb δac ) = δcb ∇n δac + δac ∇n δcb = 2 ∇n δab . So ∇n δab = 0. Problem 1.7.2 Let ∇ and ∇′ be derivative operators on a manifold, and let αa1 ...an be a smooth n−form on it. Show that

∇[b αa1 ...am ] = ∇′[b αa1 ...am ] .

a a There is a smooth, symmetric field Cbc on the manifold such that ∇′ = (∇, Cbc ). For any smooth

n−form αa1 ...an on M , we have

∇′b αa1 ...an = ∇b αa1 ...an + αr a2 ...an Cbr a1 + ... + αa1 ...an−1 r Cbr an .

273


So, anti-symmetrizing, ∇′[b αa1 ...an ] = ∇[b αa1 ...an ] + αr [a2 ...an Cbr a1 ] + ... + α[a1 ...an−1 |r| Cbr an ] . a a Since C[b c] = 0, all terms involving Cb c in the sum on the right-hand side are 0. (Notice, for example, r that αr [a2 ...an Cbr a1 ] = αr [a2 ...an C[b a1 ]] = 0.) It follows that

∇[b αa1 ...an] = ∇′[b αa1 ...an ] . Problem 1.7.3 Let ∇ be the coordinate derivative operator canonically associated with (U, ϕ) on the

n-manifold M . Let ui be the coordinate maps on U determined by the chart. Further, let ∇′ be another

a derivative operator on U . We know (from proposition 1.7.3) that there is a smooth field Cbc on U such a that ∇′ = (∇, Cbc ). Show that if a Cbc =

n X n X n X

ijk

C

i=1 j=1 k=1

∂ ∂ui

a

(db uj ) (dc uk ),

a n X i ∂ ξ then a smooth vector field ξ = on U is constant with respect to ∇′ (i.e., ∇′a ξ b = 0) iff i ∂u i=1 a

i

n X ijk k ∂ξ C ξ = j ∂u k=1

for all i and j. We have ∇′b ξ a

= =

∇b ξ a − Cbac ξ c i a a c n X n n X n X n X n X X ijk l ∂ξ ∂ ∂ ∂ j j k C ξ (db u ) − (db u )(dc u ) . ∂uj ∂ui ∂ui ∂ul i=1 j=1 i=1 j=1 k=1 l=1

c ∂ k = δk l . So, continuing, But (dc u ) ∂ul

a a n X n X n n X n X X ijk k ∂ ∂ξ ∂ j C ξ (db u ) − (db uj ) j i i ∂u ∂u ∂u i=1 j=1 k=1 i=1 j=1 " i # a n n n XX ∂ξ X ijk k ∂ C ξ − (db uj ). j i ∂u ∂u i=1 j=1 i

∇′b

ξ

a

=

=

k=1

Thus ∇′b ξ a = 0 iff every coefficient (in brackets) in the sum on the right side is 0, i.e., iff i

n

X ijk k ∂ξ C ξ − j ∂u k=1

for all i and j in {1, ..., n}.

= 0

274


a Problem 1.8.1 Let ∇ and ∇′ be derivative operators on a manifold with ∇′m = (∇m , Cbc ), and let their

a respective curvature fields be Rbcd and R′abcd . Show that

a n a a . Cd]n + 2Cb[c R′abcd = Rbcd + 2∇[c Cd]b

Given any smooth field αb , ∇′c ∇′d αb

=

a ∇′c (∇d αb + αa Cdb )

=

p p a a a ∇c (∇d αb + αa Cdb ) + (∇p αb + αa Cpb )Ccd + (∇d αp + αa Cdp )Ccb .

p Expanding the first term, anti-symmetrizing on the indices c and d, and using the fact that C[cd] = 0, we

arrive at 1 ′a 1 p p a a a R bcd αa = Rabcd αa + (∇[c α|a| )Cd]b + αa ∇[c Cd]b + (∇[d α|p| )Cc]b + αa Cb[c Cd]p . 2 2 The second and fourth terms on the right hand side differ only in their respective indices of contraction and the order in which the indices c and d occur. So their sum is 0. Hence, 1 ′a 1 p a a . + αa Cb[c Cd]p R bcd αa = Rabcd αa + αa ∇[c Cd]b 2 2 But this holds for all smooth fields αa . So our conclusion follows. Problem 1.8.2 Show that the exterior derivative operator d on any manifold satisfies d2 = 0, i.e., dn (dm αb1 ...bp ) = 0 for all smooth p-forms αb1 ...bp . Here we use the fact that λ[a...[b...c]...d] = λ[a...b...c...d] for all tensor fields λa...b...c...d. It follows from this that dn (dm αb1 ...bp )

= ∇[n ∇[m αb1 ...bp ]] = ∇[n ∇m αb1 ...bp ] = ∇[[n ∇m] αb1 ...bp ] i 1h αr [b2 ...bp Rbr1 n m] + ... + α[b1 ...bp−1 |r| Rbrp n m] = 2 i 1h r r = . + ... + α[b1 ...bp−1 |r| R[b αr [b2 ...bp R[b p n m]] 1 n m]] 2

a Since R[bcd] = 0, each of the terms in the final sum is 0. So we are done.

Problem 1.8.3 Show that given any smooth field ξ a , and any derivative operator ∇ on a manifold, £ξ

m commutes with ∇ (in its action on any tensor field) iff ∇a ∇b ξ m = Rbna ξn.

...ar m m Let Kab = Rbna ξ n − ∇a ∇b ξ m . We claim that for all smooth fields αba11...b , s ...ar (£ξ ∇n − ∇n £ξ )αba11...b s

a1 ...ar a1 ...ar m m = αm b2 ...bs Kn b1 + ... + αb1 ...bs−1 m Kn bs a ...a

a2 ...ar −αm Kna1m − ... − αb11...bsr−1 b1 ...bs

m

Knarm .

Consider first the case of a scalar field α. By proposition 1.6.4 (and the fact that ∇[m ∇n] α = 0), (£ξ ∇n − ∇n £ξ )α

=

(ξ m ∇m ∇n α + (∇m α)∇n ξ m ) − ∇n (ξ m ∇m α)

=

ξ m ∇m ∇n α + (∇m α)(∇n ξ m ) − (∇n ξ m )∇m α − ξ m ∇n ∇m α = 0.

275


Similarly, in the case of a smooth vector field αa , we have (£ξ ∇n − ∇n £ξ )αa

= [ξ m ∇m ∇n αa + (∇m αa )(∇n ξ m ) − (∇n αm )(∇m ξ a )] −∇n (ξ m ∇m αa − αm ∇m ξ a ) = [ξ m ∇m ∇n αa + (∇m αa )(∇n ξ m ) − (∇n αm )(∇m ξ a )] −[ξ m ∇n ∇m αa + (∇n ξ m )(∇m αa ) − αm ∇n ∇m ξ a − (∇n αm )(∇m ξ a )] = 2 ξ m ∇[m ∇n] αa + αm ∇n ∇m ξ a = −ξ m Rpa m n αp + αp ∇n ∇p ξ a = −αp Kna p .

The other cases now follow with a standard march through the indices. To compute (£ξ ∇n − ∇n £ξ )αb , for example, we consider an arbitrary smooth field λb and make use of our derived expressions for

(£ξ ∇n − ∇n £ξ )(αb λb ) and (£ξ ∇n − ∇n £ξ )λb . And so forth. m Now if Kab = 0, it follows immediately from our equation that £ξ commutes with ∇ in its action on

any smooth tensor field. Conversely, if the commutation condition holds, then αm Knmb = 0 for all smooth fields αb . So Knmb = 0.

Problem 1.8.4 Show that given any smooth field ξ a on a manifold, the operators £ξ and dn commute in their action on all smooth p-forms. Given any smooth p-form αb1 ...bp , we have, by the preceding problem, (£ξ dn − dn £ξ )αb1 ...bp

= £ξ ∇[n αb1 ...bp ] − ∇[n £ξ αb1 ...bp ] = αm [b2 ...bp Knmb1 ] + ... + α[b1 ...bp−1 |m| Knmbp ] m m = αm [b2 ...bp K[n b1 ]] + ... + α[b1 ...bp−1 |m| K[n bp ]] .

m Each of the terms in the final sum is 0, since K[rs] = 0. (This follows, since by the symmetries of the

Riemann tensor field, m 2 K[rs]

=

m m m m 2(R[s|n|r] ξ n − ∇[r ∇s] ξ m ) = Rsnr ξ n − Rrns ξ n + Rnrs ξn

=

m m m m ξ n = 0.) Rsnr ξ n + Rrsn ξ n + Rnrs ξ n = 3 R[snr]

So (£ξ dn − dn £ξ )αb1 ...bp = 0. Problem 1.9.1 Let ∇ be a derivative operator on a manifold that is compatible with the metric gab . Use

the Bianchi identity to show that

∇a Rab −

1 ab g R = 0. 2

By the Bianchi identity, and various symmetries of the Riemann tensor field, we have 0 = ∇m Rabcd + ∇d Rabmc + ∇c Rabdm = ∇m Rabcd − ∇d Rbamc − ∇c Rabmd .

276


If we raise indices a and b, and then perform (a, d) and (b, c) contraction, we arrive at 0 = ∇m R − ∇a Ram − ∇b Rbm . Contracting with g mc (and changing indices of contraction) yields 0 = ∇m (g mc R) − 2 ∇a Rac = ∇a (g ac R − 2 Rac ). So, ∇a Rac −

1 2

g ac R = 0.

Problem 1.9.2 Let ξ a be a smooth vector field on M . Show that £ξ g ab = 0 ⇐⇒ £ξ gab = 0. We know that £ξ δ ac = 0 (Problem 1.6.1). Hence 0 = £ξ δ ac = £ξ (g ab gbc ) = g ab £ξ gbc + gbc £ξ g ab . Assume that £ξ gab = 0. Then gbc £ξ g ab = 0 and, therefore, 0 = g cd gbc £ξ g ab = δb d £ξ g ab = £ξ g ad . This gives us the implication from left to right The converse is handled similarly. Problem 1.9.3 Show that Killing fields on M with respect to gab are affine collineations with respect to ∇. Let ξ a be a Killing field. By proposition 1.9.8 (and various symmetries of the Riemann curvature tensor), ∇a ∇b ξm = −Rnabm ξn = −Rnabm ξ n = −Rbmna ξ n = Rmbna ξ n . So ∇a ∇b ξ m = Rmbna ξ n . It now follows immediately from problem 1.8.3 that ξ a is an affine collineation

with respect to ∇.

Problem 1.9.4 Show that if ξ a is a Killing field on M with respect to gab , then the Lie derivative operator £ξ annihilates the fields Rabcd , Rab , and R determined by gab . Given any smooth vector field η a , we have £ξ (Rabcd ηa ) = £ξ (2 ∇[c ∇d] ηb ) = 2 ∇[c ∇d] (£ξ ηb ) = Rabcd £ξ ηa . (The second equality follows from the preceding problem. Since ξ a is a Killing field, it is an affine collineation with respect to ∇, i.e., £ξ commutes with ∇.) But by the Leibniz rule, we also have £ξ (Rabcd ηa ) = Rabcd £ξ ηa + ηa £ξ Rabcd .

277


Comparing these two expressions, we see that ηa £ξ Rabcd = 0. But this is true for all smooth fields ηa . So £ξ Rabcd = 0. Hence, since £ξ δ nm = 0, £ξ Rab = £ξ (δ nm Rmabn ) = δ nm £ξ Rmabn = 0. Since ξ a is a Killing field, £ξ g mn = 0. (See problem 1.9.2.) So it follows that £ξ R = £ξ (g ab Rab ) = g ab £ξ Rab = 0.

Problem 1.9.5 Show that if ξ a and η a are Killing fields on M with respect to gab , and k is a real number, then (ξ a + η a ), (k ξ a ), and the commutator [ξ, η]a = £ξ η a are all Killing fields with respect to gab as well. λa = (ξ a + η a ) is a Killing field since ∇(a λb) = ∇(a ξb) + ∇(a ηb) = 0. Similarly, χa = (kξ a ) is a Killing field since ∇(a χb) = k ∇(a ξb) = 0. Finally, θa = £ξ η a is a Killing field since, by problem 1.6.6, £θ gab = £ξ £η gab − £η £ξ gab = 0. Problem 1.9.6 Let η a be a Killing field on M with respect to gab . (i) Let γ be a geodesic with tangent field ξ a . Show that the function E = ξ a ηa is constant on γ. (ii) Let T ab be a smooth tensor field that is symmetric and divergence free (i.e., ∇a T ab = 0), and let J a be the field T ab ηb . Show that ∇a J a = 0. Let η a , γ, ξ a , and E be as stated. Then we have ξ n ∇n E = ξ n ∇n (ξ a ηa ) = ξ n ξ a ∇n ηa + ηa ξ n ∇n ξ a . Since ξ a is a Killing field, ∇n ηa is anti-symmetric. So ξ n ξ a ∇n ηa = 0. And since ξ a is the tangent field of

a geodesic, ξ n ∇n ξ a = 0. So, ξ n ∇n E = 0. This gives us (1). The computation for (2) is much the same: ∇a J a = ∇a (T ab ηb ) = T ab ∇a ηb + ηb ∇a T ab .

The second term on the right side vanishes since ∇a T ab = 0. The first vanishes since T ab is symmetric

(and hence T ab ∇a ηb = T ab ∇(a ηb) = 0). So ∇a J a is 0.

Problem 1.9.7 Show that if η a is a conformal Killing field on M , and M has dimension n, then ∇(a ηb) =

1 (∇c η c ) gab . n

Assume £η Ω2 gab = 0. Then, by proposition 1.7.4 (and the fact that ∇m gab = 0), 0 = =

Ω2 £η gab + gab £η Ω2 = Ω2 [η m ∇m gab + ∇a ηb + ∇b ηa ] + gab £η Ω2

Ω2 [∇a ηb + ∇b ηa ] + gab £η Ω2 .

278


If we raise the index ‘b’ and then contract, we obtain 0 = 2 Ω2 (∇a η a ) + n£η Ω2 . Our two equations jointly yield ∇(a ηb) =

1 gab (∇c η c ). n

Problem 1.10.1 Let S be a k-dimensional imbedded submanifold of the n-dimensional manifold M , and let p be a point in S. (1) Show that the space of co-vectors ηa ∈ (Mp )a normal to S has dimension (n − k). (2) Show that a vector ξ a ∈ (Mp )a is tangent to S iff ηa ξ a = 0 for all co-vectors ηa ∈ (Mp )a that are normal to S.

1

2

k

(1) The subspace of vectors in (Mp )a tangent to S has dimension k. Let {ξ a , ξ a , ..., ξ a } be any set

of k linearly independent vectors from that subspace. We can extend it to a basis for (Mp )a by adding k+1

n

1

n

(n − k) more (appropriately chosen) vectors ξ a , ..., ξ a . Now let {α a , ..., α a } be the dual basis. So i

j

k+1

n

α a ξ a = δij . We claim that the subspace of co-vectors at p normal to S is spanned by { α a , ..., α a }. To 1 1

n n

i

i

see this, consider any co-vector αa = α α a + ...+ α α a at p. It is normal to S iff α = αa ξ a = 0 for 1

2

k

all i = 1, ..., k (since every vector at p tangent to S is a linear combination of ξ a , ξ a , ..., ξ a ). Thus α is k+1

n

normal iff it is in the linear span of { α a , ..., α a }. So the latter is a basis for the subspace of co-vectors

at p normal to S — and therefore that subspace has dimension (n − k).

(2) The argument is much the same. We continue to work with the basis and dual basis described in 11

nn

(1).. Consider any vector ξ a = ξ ξ a + ...+ ξ ξ a at p. It is killed by every covariant vector at p normal k+1

n

i

i

to S iff it is killed by all the vectors α a , ..., α a . And the latter condition holds iff ξ = αa ξ a = 0 for all i = k + 1, ..., n. So ξ a is killed by every covariant vector at p normal to S iff it is a linear combination of 1

2

k

ξ a , ξ a , ..., ξ a , i.e., iff it is tangent to S. Problem 1.10.2 Let S be a k-dimensional imbedded submanifold of the n-dimensional manifold M , and let gab be a metric on M . Show that S is a metric submanifold (relative to gab ) iff, for all p in S, the pull-back tensor (idp )∗ (gab ) is non-degenerate, i.e., there is no non-zero vector ξã ∈ (Sp )a such that (idp )∗ (gab ) ξã = 0.

Let p be any point in S. The pull-back tensor (idp )∗ (gab ) is degenerate there iff there is a ξã ∈ (Sp )a η b ) = 0. Since a vector in such that, for all ηã ∈ (Sp )a , (idp )∗ (gab ) ξã η˜b = gab (idp )∗ (ξã ) (idp )∗ (˜

(Mp )a is tangent to S precisely if it is of the form (idp )∗ (ηã ) for some ηã ∈ (Sp )a , we see that (idp )∗ (gab ) is degenerate at p iff there is a ξã ∈ (Sp )a such that gab (idp )∗ (ξ˜b ) is normal to S, i.e., there is a vector

in (Mp )a tangent to S that is also normal to S.

Problem 1.10.3 Prove the following generalization of clause (2) in proposition 1.10.3. For all M -tensor fields α ...a... on S, the following conditions both hold.

279


(1) α ...a... is tangent to S in the index ‘a’ ⇐⇒ hab α ...b... = α ...a... ⇐⇒ k ab α ...b... = 0. (2) α ...a... is normal to S in the index ‘a’ ⇐⇒ k ab α ...b... = α ...a... ⇐⇒ hab α ...b... = 0.

We work with a representative case. (The proof is exactly the same no matter how many indices are involved.) Consider the M -field αamn on S. Suppose first that hab αbmn = αamn . Then αamn is certainly tangent to S in a since hab is. Conversely, suppose αamn is tangent to S in a. Then, we claim, hab αbmn and αamn have the same action on any co-vector ηa (at any point of S) that is either tangent to, or normal to, S. In the first case, hab αbmn ηa = αamn ηa , since hab ηa = ηb , In the second case, hab αbmn ηa = 0 = αamn ηa , because both sides are tangent to S in ‘a’. This gives us the first equivalence in (1). The second is immediate since k ab αbmn = (g ab − hab ) αbmn = αamn − hab αbmn . The equivalences

in (2) are handled similarly.

Problem 1.10.4 Prove that hma k nb k pc ∇m hnp = 0. We have hma k nb k p c ∇m hnp = hma k nb [∇m (hnp k pc ) − hnp ∇m k pc ]. But hnp k pc = 0 = k nb hnp (by the third clause of proposition 1.10.3). So both terms on the right are 0. Problem 1.10.5 Derive the second Gauss-Codazzi equation: hm[a hnb] hpc k rd ∇m πnpr =

1 m n p r h h h k Rmnpr . 2 a b c d

We have (1) hma hnb hpc ∇m hnp = 0. (2) hma k nb k pc ∇m hnp = 0. (3) πabc = hma hnb k pc ∇m hnp = hma hnb (g pc − hpc ) ∇m hnp = hma hnb ∇m hnc . (The first is (1.10.3); the second was proved in the preceding problem; and the third follows from the first.) Hence hm[a hnb] hpc k rd ∇m πnpr = hm[a hnb] hpc k rd ∇m (hqn hsp ∇q hsr ) = A + B + C where A =

hm[a hnb] hpc k rd (∇m hqn ) hsp (∇q hsr ),

B

=

hm[a hnb] hpc k rd hqn (∇m hsp ) (∇q hsr ),

C

=

hm[a hnb] hpc k rd hqn hsp (∇m ∇q hsr ).

By (3) and lemma 1.10.6, hm[a hnb] (∇m hqn ) = π[ab]q = 0. So A = 0. Furthermore, since hnb hqn = hqb , we have, by (3), B

=

hm[a hqb] hpc k rd (∇m hsp ) (∇q hsr ) = hq[b hma] hpc (∇m hsp ) k rd (∇q hsr )

=

hq[b πa]cs k rd (∇q hsr ).

280


Now πacs is tangent to S in the index a and normal to it in s. So πac s = hua k sv πuc v and therefore, continuing the computation, B = hq[b πa]cs k rd (∇q hsr ) = hq[b hua] k sv k rd (∇q hsr )πuc v = 0. (The final equality follows from (2).) Finally, since hnb hqn = hqb , C

= hm[a hqb] hpc k rd hsp (∇m ∇q hsr ) = hma hqb hpc k rd hsp (∇[m ∇q] hsr ) =

1 m q p r s h a h b h c k d h p (hur Rusmq + hsu Rurmq ). 2

Now k rd hur = 0 (by proposition 1.10.3) and hsp hsu = hpu . So, continuing, hm[a hnb] hpc k rd ∇m πnpr

1 m q p r 1 h a h b h c k d hpu Ru rmq = hma hqb hpc k rd (gpu − kpu ) Ru rmq 2 2

=

C =

=

1 m q p r h a h b h c k d Rprmq . 2

(The final equality follows from the fact that, once again, hpc kpu = 0.) But Rprmq = Rmqpr . So we are done. Problem 1.11.1 One learns in the study of ordinary vector analysis that, for all vectors ξ, η, θ, λ at a point, the following identities hold. (1) (ξ × η) · (θ × λ) = (ξ · θ)(η · λ) − (ξ · λ)(η · θ) (2) (ξ × (η × θ)) + (θ × (ξ × η)) + (η × (θ × ξ)) = 0. Reformulate these assertions in our notation and prove them. The two come out as follows. (1′ ) (ǫabc ξb ηc )(ǫamn θm λn ) = (ξ b θb )(η c λc ) − (ξ b λb )(η c θc ). (2′ ) ǫabc ξb (ǫcmn η m θn ) + ǫabc θb (ǫcmn ξ m η n ) + ǫabc ηb (ǫcmn θm ξ n ) = 0. They follow easily from (1.11.6) — in the case where n = 3 and n− = 0. First, we have (ǫabc ξb ηc )(ǫamn θm λn )

= (ǫabc ǫamn )ξb ηc θm λn = 2 δ [bm δ c]n ξb ηc θm λn = 2 ξb ηc θ[b λc] = (ξ b θb )(η c λc ) − (ξ b λb )(η c θc ).

And for the second, we have ǫabc ξb (ǫcmn η m θn ) +

ǫabc θb (ǫcmn ξ m η n ) + ǫabc ηb (ǫcmn θm ξ n )

=

(ǫcab ǫcmn )ξb η m θn + (ǫcab ǫcmn )θb ξ m η n + (ǫcab ǫcmn )ηb θm ξ n

=

2 δ [am δ b]n ξb η m θn + 2 δ [am δ b]n θb ξ m η n + 2 δ [am δ b]n ηb θm ξ n

=

2 ξb η [a θb] + 2 θb ξ [a η b] + 2 ηb θ[a ξ b]

=

[(ξb θb )η a − (ξb η b )θa ] + [(θb η b )ξ a − (θb ξ b )η a ] + [(ηb ξ b )θa − (ηb θb )ξ a ]

=

0.

281


Problem 1.11.2 Do the same for the following assertion: div(ξ × η) = η · curl(ξ) − ξ · curl(η). We have ∇a (ǫabc ξb ηc ) = =

ξb ǫabc ∇a ηc + ηc ǫabc ∇a ξb ηc ǫcab ∇a ξb − ξb ǫbac ∇a ηc .

Problem 1.11.3 We have seen that every Killing field ξ a in n-dimensional Euclidean space (n ≥ 1) can be expressed uniquely in the form

ξb = χa Fab + kb , where Fab and kb are constant, Fab is anti-symmetric, and χa is the position field relative to some point p. Consider the special case where n = 3. Let ǫabc be a volume element. Show that (in this special case) there is a unique constant field W a such that Fab = ǫabc W c . Let W a = 12 ǫabc Fbc . Then 1 1 n] ǫabc W c = ǫabc ( ǫcmn Fmn ) = (ǫcmn ǫcab )Fmn = δ [ma δ b Fmn = F[ab] = Fab . 2 2 (The final equality follows from the fact that Fab is anti-symmetric.) W a is constant, since ∇b W a =

1 ∇b (ǫamn Fmn ), 2

and both ǫamn and Fmn are constant. Finally, W a is the unique field satisfying the given constraint, for c c , then ǫabc (W c c − W c ) = 0, and so if we also have Fab = ǫabc W c c − W c ) = 2 δ n (W c c − W c ) = 2 (W c n − W n ). 0 = ǫabn ǫabc (W c

Problem 2.1.1 Consider our characterization of timelike vectors in terms of null vectors in the proof of proposition 2.1.1. Why does it fail if n = 2? If n = 2, the stated condition holds for spacelike as well as timelike vectors. Indeed, in that dimension, given any two non-zero null vectors αa and γ a that are not proportional to one another, every spacelike (as well as every timelike) vector η a can be expressed in the form η a = kαa + lγ a , where k 6= 0 and l 6= 0. So, of course, if we take β a to be the null vector lγ a , then we have η a = kαa + β a .

Problem 2.1.2 (i) Show that it is possible to characterize timelike vectors in terms of causal vectors. (ii) Show that it is possible to characterize timelike vectors in terms of spacelike vectors. The following equivalences hold for all n ≥ 2. A vector η a at p is timelike iff for all causal vectors αa at p, there is an ǫ > 0 such that, for all k, if |k| < ǫ, then η a + kαa is causal.

282


A vector η a at p is timelike iff for all spacelike vectors αa at p, there is an ǫ > 0 such that, for all k, if |k| < ǫ, then η a + kαa is not spacelike. Problem 2.1.3 Does proposition 2.1.3 still hold if condition (1) is left intact but (2) is replaced by ...ar b1 (2 ′ ) αba11...b ξ ... ξ bs = 0 for all spacelike vectors ξ a at the point? s

And what if it is replaced by ...ar b1 (2 ′′ ) αba11...b ξ ... ξ bs = 0 for all null vectors ξ a at the point? s

Condition (2 ′′ ) is certainly not sufficient. For example, if gab is a spacetime metric and p is a point in the underlying manifold, then gab ξ a ξ b = 0 for all null vectors ξ a at p, but gab 6= 0. On the other hand,

condition (2 ′ ) is sufficient, and the proof is almost the same as for the original version of proposition 2.1.3. Only one change is needed. Before we used the fact that if ξ a is a timelike vector at some point, and η a is an arbitrary vector there, then there is an ǫ > 0 such that, for all x, if |x| < ǫ, then (ξ a + xη a ) is timelike. Now we use the corresponding assertion with both occurences of “timelike” changed to “spacelike”. Problem 2.2.1 Let p be a point in M . Show that there is no two-dimensional subspace of Mp all of whose elements are causal (timelike or null). Assume there are non-zero, linearly independent vectors αa and β a at p such that, for all k and l, the vector (k αa + l β a ) is causal. We derive a contradiction. There are two cases to consider. Either (i) one of the two is timelike, or (ii) both are null. Assume first that one of the two, say αa , is timelike. If we set k=−

(αm β m ) (αn αn )

l = 1,

then αn (k αn + l βn ) = 0, i.e., (k αa + l β a ) is orthogonal to the timelike vector αa . Since (k αa + l β a ) is causal, it follows from the first clause of proposition 2.2.1 that (k αa + l β a ) = 0. This contradicts our assumption that αa and β a are linearly independent. Assume next that αa and β a are both null. Then, for all k and l, 0 ≤ (k αn + l β n )(k αn + l βn ) = 2 k l (αn βn ), since (k αa + l β a ) is causal. But this can hold for all k and l only if (αn βn ) = 0. Hence, by the second clause of proposition 2.2.1, αa and β a must be proportional to one another. Once again, this contradicts our assumption that they are linearly independent. Problem 2.2.2 Let g ′ab be a second metric on M (not necessarily of Lorentz signature). Show that the following conditions are equivalent. (1) For all p in M , gab and g ′ab agree on which vectors at p are orthogonal.

283


(2) g ′ab is conformally equivalent to either gab or −gab . The implication (2) ⇒ (1) is immediate. For the other direction, assume (1) holds. It follows from (1)

that gab and g ′ab agree as to which vectors are null, i.e., orthogonal to themselves. So it will suffice to show that g ′ab has signature (1, 3) or (3, 1). For then we can invoke proposition 2.1.1 and conclude that g ′ab is conformally equivalent to gab (in the first case) or to −gab (in the second case). 1

4

Let p be any point in M , and let ξa , ..., ξa be an orthonormal basis at p with respect to gab . Consider 1

2

the vector (ξ a + ξ a ). It is null with respect to gab . So it must be null with respect to g ′ab . Furthermore, 1

2

since ξ a and ξ a are orthogonal with respect to gab , they must be orthogonal with respect to g ′ab . So we have 0

1

2

1

1

2

1

2

2

= g ′ab (ξ a + ξ a )(ξ b + ξ b ) = g ′ab ξ a ξ b + g ′ab ξ a ξ b .

Similarly, we have

i

1

1

3

3

1

1

4

4

0 =

g ′ab ξ a ξ b + g ′ab ξ a ξ b

0 =

g ′ab ξ a ξ b + g ′ab ξ a ξ b .

i

i

Now let Xi = g ′ab ξ a ξ b , for i = 1, ..., 4. The Xi are non-zero — since the vectors ξ a are non-null with respect to gab . So there are only two possibilities. Either X1 > 0 and X2 , X3 , X4 < 0, or X1 < 0 and X2 , X3 , X4 > 0. In the first case, g ′ab has signature (1, 3); in the second, it has signature (3, 1). (In 1

4

either case, we need only normalize the vectors ξ a , ..., ξ a to arrive at an orthonormal basis at p of the appropriate type for g ′ab .) Problem 2.2.3 Prove the second clause of proposition 2.2.3. Let µa and ν a be co-oriented, non-zero causal vectors at a point p. Then either (µn νn ) > 0, or both vectors are null and µa = k ν a for some k > 0. In the latter case, kµa + ν a k = kµa k = kν a k = 0, and the

assertion follows trivially. So we may assume (µn νn ) > 0. Hence, by the first clause of proposition 2.2.3, (µn νn ) ≥ kµa k kν a k. Therefore, 2

(kµa k + kν a k)

=

kµa k2 + 2 kµa k kν a k + kν a k2 ≤ (µn µn ) + 2 (µn νn ) + (ν n νn )

=

(µn + ν n ) (µn + νn ) = kµa + ν a k2 .

(For the final equality we need the fact µa and ν a are co-oriented. Otherwise, (µa + ν a ) need not be causal.) Equality holds here iff (µn νn ) = kµa k kν a k. But by the first half of the proposition, again, this

is the case iff µa and ν a are proportional.

Problem 2.5.1 Give examples for each of the following possibilities. (1) A smooth symmetric field Tab that does not satisfy the WEC. (2) A smooth symmetric field Tab that satisfies the WEC but not the DEC.

284


(3) A smooth symmetric field Tab that satisfies the DEC but not the SDEC. (1) Tab = −gab . (2) Tab = σa σb , where σ a is a smooth spacelike field. (3) Tab = λa λb , where λa is a

smooth, non-zero null field.

Problem 2.5.2 Show that the DEC holds iff given any two co-oriented timelike vectors ξ a and η a at a point, Tab ξ a η b ≥ 0. Suppose first that the DEC holds, and let ξ a be a timelike vector at some point. Then Tab ξ a ξ b ≥ 0

and T a b ξ b is a causal vector. We claim that Tab ξ a η b ≥ 0 for all timelike η a at the point that are co-

oriented with ξ a . We may assume that Tab ξ a 6= 0, since otherwise the claim is trivial. And in this case

it follows that Tab ξ a ξ b > 0 (since otherwise Tab ξ a is a non-zero causal vector that is orthogonal to the

timelike vector ξ b , which is impossible by proposition 2.2.1). So T a b ξ b is a non-zero causal vector that is co-oriented with ξ a . Now let η a be any timelike vector at the point that is co-oriented with ξ a . It must be co-oriented with T a b ξ b as well (since co-orientation is an equivalence relation). So Tab ξ a η b > 0. For the converse, suppose that given any two co-oriented timelike vectors ξ a and η a at a point, Tab ξ a η b ≥ 0. Let ξ a be a timelike vector at some point. It follows immediately (taking η a = ξ a ) that

Tab ξ a ξ b ≥ 0. So what we have to show is that T a b ξ b is a causal vector. Suppose to the contrary that it o

is spacelike. Then we can find a timelike vector η a at the point, co-oriented with ξ a , that is orthogonal to o

o

T a b ξ b . But since η a is timelike, (η a + k T a b ξ b ) is also timelike and co-oriented with ξ a for all sufficiently small k > 0. Hence, by our initial assumption, o

0 ≤ Tab ξ a (η b + k T b n ξ n ) = k (Tab ξ a ) (T b n ξ n ) for all sufficiently small k > 0. But this is impossible since (Tab ξ a )(T b n ξ n ) < 0. So, as claimed, T a b ξ b is causal. Problem 2.5.3

Consider a perfect fluid with four-velocity η a , energy density ρ, and pressure p.

(i) Show that it satisfies the DEC iff DEC.

|p| ≤ ρ. (ii) Show that it satisfies the SDEC iff it satisfies the

(i) Suppose Tab = ρ ηa ηb − p (gab − ηa ηb ). Then Tab satisfies the DEC condition at a point iff for all

unit timelike vectors ξ a at that point, Tab ξ a ξ b ≥ 0 and T a b ξ b is causal. Now for all such vectors Tab ξ a ξ b

=

(T a b ξ b )(Tac ξ c ) =

(ρ + p)(η a ξa )2 − p, (ρ2 − p2 )(η a ξa )2 + p2 .

So the DEC holds iff both right-side expressions are non-negative for all choices of ξ a . Assume first that |p| ≤ ρ, and let ξ a be a unit timelike vector at the point in question. Then, by the

wrong-way Schwarz inequality (proposition 2.2.3), (η a ξa )2 ≥ kη a k2 kξ a k2 = 1. Hence, (ρ + p)(η a ξa )2 − p (ρ2 − p2 )(η a ξa )2 + p2

≥ (ρ + p) − p = ρ ≥ 0, ≥ (ρ2 − p2 ) + p2 = ρ2 ≥ 0.

285


So the DEC holds at the point. Conversely, suppose that Tab ξ a ξ b ≥ 0 and T a b ξ b is causal for all unit

timelike vectors ξ a at the point. Then, in particular, Tab η a η b ≥ 0 and, therefore, 0 ≤ (ρ+p)(η a ηa )2 −p =

(ρ+p)−p = ρ there. Next we use the fact that there is no upper bound to the value of (η a ξa )2 as ξ a ranges over unit timelike vectors at the point. It cannot possibly be the case that (ρ2 − p2 )(η a ξa )2 + p2 ≥ 0

for all such vectors unless (ρ2 − p2 ) ≥ 0. So we have ρ ≥ 0 and (ρ2 − p2 ) ≥ 0. These two together are jointly equivalent to |p| ≤ ρ, as required.

(ii) The SDEC implies the DEC (always, not just for perfect fluids). Suppose that at some point Tab = ρ ηa ηb − p (gab − ηa ηb ) satisfies the DEC but not the SDEC. Then there is a timelike vector ξ a

at the point such that T a b ξ b is null even though Tab 6= 0 there. We claim this is impossible. If T a b ξ b

is null, then, (ρ2 − p2 )(η a ξa )2 + p2 = 0. But |p| ≤ ρ, since we are assuming that the DEC holds, and

η a ξa 6= 0 (since no two timelike vectors are orthogonal). So this equation can hold only if ρ = p = 0, and

this contradicts our assumption that Tab 6= 0 at the point in question.

Problem 2.6.1 Show that Maxwell’s equations in the source free case (J a = 0) are conformally invariant. ′ = Ω2 gab be a second metric on the underlying manifold M , whose dimension n we leave open. Let gab

Let its associated derivative operator be ∇′ . It will suffice for us to show that ∇′a (g ′am g ′bn Fmn ) =

1 (n − 4) ab (∇a F ab ) + F ∇a Ω. Ω4 Ω5

We know from proposition 1.9.5 that ∇′ = (∇, C a bc ), where C abc = −

i 1 h a 2 a 2 ar 2 . δ ∇ Ω + δ ∇ Ω − g g ∇ Ω c b bc r c 2 Ω2 b

We have ∇′a (g ′am g ′bn Fmn ) = =

g ′am g ′bn ∇′a Fmn = Ω−4 g am g bn ∇′a Fmn Ω−4 g am g bn [∇a Fmn + C r am Frn + C r an Fmr ] .

Now Ω−4 g am g bn C r am Frn

i 1 am bn h r δ a ∇m Ω2 + δ rm ∇a Ω2 − gam g rs ∇s Ω2 Frn g g 6 2Ω i 1 h mb = − F ∇m Ω2 + F ab ∇a Ω2 − n F sb ∇s Ω2 6 2Ω (n − 2) ab F ∇a Ω = Ω5 = −

and (since Fab is anti-symmetric and, therefore, g rm Fmr = 0), Ω−4 g am g bn C r an Fmr

i 1 am bn h r 2 r 2 rs 2 g g δ ∇ Ω + δ ∇ Ω − g g ∇ Ω Fmr a n a an s n 2 Ω6 i h 1 g rm Fmr g bn ∇n Ω2 + F ab ∇a Ω2 − F bs ∇s Ω2 = − 2 Ω6 2 = − 5 F ab ∇a Ω. Ω

= −

286


So, as needed, we have ∇′a (g ′am g ′bn Fmn ) =

1 (n − 4) ab (∇a F ab ) + F ∇a Ω. Ω4 Ω5

Problem 2.6.2 Prove (2.6.19). We have

ǫabcd Fab Fcd = ǫabcd 2 E[a ξb] + ǫabrs ξ r B s 2 E[c ξd] + ǫcdmn ξ m B n .

When we expand the right side, we get four terms. Two of them vanish because of the anti-symmetry of ǫabcd : ǫabcd E[a ξb] E[c ξd]

=

ǫabcd Ea ξb Ec ξd = 0,

ǫabcd ǫabrs ξ r B s ǫcdmn ξ m B n

=

− 4 δ c[r δ d s] ξ r B s ǫcdmn ξ m B n = − 4 ξ c B d ǫcdmn ξ m B n = 0.

One of the other terms yields 2 ǫabcd E[a ξb] ǫcdmn ξ m B n = 2 ǫcdab ǫcdmn Ea ξb ξ m B n = − 8 δ a [m δ b n] Ea ξb ξ m B n = 4 E a Ba , since ξ a Ea = ξ a Ba = 0. The other yields 4 E a Ba as well. (The computation is almost exactly the same.) So we have ǫabcd Fab Fcd = 8 E a Ba .

Problem 2.6.3 Prove (2.6.21). By (2.6.17), we have 1 1 (Tab ξ b )(T ac ξc ) = (−E n En − B n Bn ) ξa − ǫars E r B s (−E m Em − B m Bm ) ξ a − ǫapq Ep Bq . 2 2 The two “cross-terms” on the right vanish since ξa ǫapq = ξa ǫapqn ξn = 0. So (Tab ξ b )(T ac ξc ) =

1 n (E En + B n Bn )2 + ǫars E r B s ǫapq Ep Bq . 4

But ǫars E r B s ǫapq Ep Bq

= ǫarsn ξ n ǫapqm ξm E r B s Ep Bq = −6 E [p B q ξ m] Ep Bq ξm = − (E p Ep )(B q Bq ) − (E p Bp )2 .

So we may conclude, as required, that

(Tab ξ b )(T ac ξc ) =

1 n (E En − B n Bn )2 + (E n Bn )2 . 4

Problem 2.6.4 Prove the following equivalence.   D Eb b ∇a F ab = J b ⇐⇒  ǫabc Db Bc

=

µ

=

ξ b ∇b E a + j a .

287


Clearly, (∇a F ab − J b ) vanishes iff its projections tangent to, and orthogonal to, ξ b both vanish, i.e.,   ξ (∇ F ab − J b ) = 0 b a ∇a F ab = J b ⇐⇒  hc b (∇a F ab − J b ) = 0.

We shall work on the right-side equations separately. Since ξ a (and, therefore, hab ) are constant, and since E a is orthogonal to ξ a , ξb (∇a F ab − J b ) = =

∇a (F ab ξb ) − (J b ξb ) = ∇a E a − µ = ∇a (ha n hn m E m ) − µ ha b hb m ∇a E m − µ = Db E b − µ.

This gives us the first equivalence. The second is handled similarly using (2.6.12) and (2.6.13). We have hc b (∇a F ab − J b ) = ∇a (F ab hc b ) − (J b hc b ) = ∇a (F ab hc b ) − j c and

i h ∇a (F ab hc b ) = ∇a (2 E [a ξ b] + ǫabrs ξr Bs ) hc b = − ξ a ∇a E c + ǫabrs ξr hc b ∇a Bs = − ξ a ∇a E c + ǫcas ∇a Bs = − ξ a ∇a E c + (ǫcmn ha m hs n ) ∇a Bs

= − ξ a ∇a E c + ǫcmn Dm Bn . So hc b (∇a F ab − J b ) = 0

⇐⇒

ǫcmn Dm Bn = ξ a ∇a E c + j c .

Problem 2.7.1 Show that in the general case (n ≥ 3), inversion of (2.7.3) leads to 1 2 Rab = 8 π Tab − T gab − λ gab . (n − 2) (n − 2) Contraction of Rab −

1 R gab − Λ gab = 8 π Tab 2

yields R− or, equivalently,

1 R n − Λ n = 8 π T, 2

(2 − n) R = 8 π T + n Λ. 2

So, substitution for R in the first equation yields Rab

= =

1 1 (8 π T + n Λ) gab + Λ gab 8 π Tab + R gab + Λ gab = 8 π Tab + 2 (2 − n) 1 2 8 π Tab − T gab − λ gab , (n − 2) (n − 2)

as required. Problem 2.7.2 Give examples of the following.

288


(1) A smooth symmetric field Tab that satisfies the SDEC (and so also the WEC and DEC) but not the strong energy condition (2) A smooth symmetric field Tab that satisfies the strong energy condition but not the WEC (and so not the DEC or SDEC either) For (1), take Tab = gab . It satisfies the SDEC. But in this case, Tab − 21 gab T = −gab , and so it does

not satisfy the SEC.

For (2), take Tab = −gab . It does not satisfy the WEC. But in this case, Tab − 12 gab T = gab , so it does

satisfy the SEC.

Problem 2.7.3 Consider a perfect fluid with four-velocity η a , energy density ρ, and pressure p. Show that it satisfies the strong energy condition iff (ρ + p) ≥ 0 and (ρ + 3p) ≥ 0. If Tab = ρ ηa ηb − p (gab − ηa ηb ), then T = (ρ − 3 p), and It follows that Tab

1 (p − ρ) gab T = (ρ + p) ηa ηb + gab . 2 2 satisfies the SEC iff, given any unit timelike vector ξ a at any point, Tab −

(ρ + p)(ηa ξ a )2 +

(p − ρ) ≥ 0. 2

Now (ηa ξ a )2 ≥ 1 by the Schwarz inequality. So if (ρ + p) ≥ 0 and (ρ + 3p) ≥ 0, then (ρ + p)(ηa ξ a )2 +

(p − ρ) (ρ + 3p) (p − ρ) ≥ (ρ + p) + = ≥ 0, 2 2 2

and the inequality is satisfied. Conversely, suppose it is satisfied for all unit timelike vectors ξ a at some point. Then, in particular, it is satisfied for ξ a = η a , which yields (ρ + 3p) ≥ 0. And since (ηa ξ a )2 can

assume arbitrarily large values as ξ a ranges over all unit timelike vectors at a point, it must be the case that (ρ + p) ≥ 0. Problem 2.8.1 Prove (2.8.8). It follows from the definition (2.8.6) of the twist vector that ǫabcd ξ c ω d =

1 ǫabcd ξ c ǫdmnr ξm ωnr = 3 δ m [a δ n b δ r c] ξ c ξm ωnr = 3 ξ c ξ[a ωbc] = ωab . 2

For the final equality, we use the fact that ξ a is orthogonal to ωab in both indices (and ωab is antisymmetric). Problem 2.8.2 Show that, at any point, ω a = 0 iff ξ[a ∇b ξc] = 0. We know from (2.8.7) and the anti-symmetry of ǫabcd that ωa =

1 abcd 1 abcd ǫ ξb ∇c ξd = ǫ ξ[b ∇c ξd] . 2 2

289


So the “if” half of the equivalence follows immediately. For the other direction, assume that ω a = 0 holds at some point. Then at that point we have 0 = ǫamnr ω a =

1 ǫamnr ǫabcd ξ[b ∇c ξd] = − 3 δ b [m δ c n δ d r] ξ[b ∇c ξd] = − 3 ξ[m ∇n ξr] . 2

Problem 2.8.3 Complete the calculation in (2.8.10). We have to compute ωb a η b ωca η c . ρn ρn We work separately with the numerator and denominator. It follows, first, from (2.8.8) that ωb a η b ωca η c

=

(ǫb a mn ξ m ω n ) η b (ǫcars ξ r ω s ) η c = ǫabmn ηb ξm ωn ǫacrs η c ξ r ω s

=

− 6 δ b [c δ m r δ n s] ηb ξm ωn η c ξ r ω s = − 6 η[c ξr ωs] η c ξ r ω s − (ηc η c )(ωs ω s ) − (ηs ω s )2 .

=

(For the final equality, we use the fact that we are doing the computation at the “initial point” where η a is orthogonal to ξ a .) And, since ρn = η n −

η b ωb n ω , ω m ωm

we have n

ρ ρn

= =

So,

η b ωb n (η c ωc )2 η c ωc (η c ωc )2 η − m ηn − r ω ωn = (η n ηn ) − 2 r + ω ωm ω ωr ω ωr ω r ωr n 1 (η ηn )(ω s ωs ) − (η c ωc )2 . (ω r ωr ) n

ωb a η b ωca η c = − ω r ωr , ρn ρn as required. Problem 2.9.1 Let κa be a timelike Killing field that is locally hypersurface orthogonal (κ[a ∇b κc] = 0). Further, let κ be the length of κa . (So κ2 = κn κn .) Show that

κ2 ∇a κb = − κ[a ∇b] κ2 . This follows with a simple direct computation: 0

= 3 κc κ[a ∇b κc] = κa κc ∇b κc + (κc κc ) ∇a κb + κb κc ∇c κa = κa κc ∇b κc + (κc κc ) ∇a κb − κb κc ∇a κc 1 1 κa ∇b κ2 + κ2 ∇a κb − κb ∇a κ2 = κ2 ∇a κb + κ[a ∇b] κ2 . = 2 2

290


Problem 2.9.2 Consider a non-trivial boost Killing field κb = 2χa E[a ξb] on Minkowksi spacetime (as determined relative to some point p and some constant unit timelike field ξ a ). “Non-trivial” here means that E a 6= 0. Let η a be a constant field on Minkowski spacetime. Show that £κ η a = 0 iff η a is orthogonal

to both to ξ a and E a .

Since η a is constant £κ η a

= κn ∇n η a − η n ∇n κa = − η n ∇n κa = − 2 η n ∇n (χm E [m ξ a] ) = − 2 E [m ξ a] η n ∇n χm = − 2 E [m ξ a] η n gnm = (ξ m ηm )E a − (E m ηm )ξ a .

Since ξ a and E a are linearly independent, we see that £κ η a = 0 iff (ξ m ηm ) = 0 = (E m ηm ). Problem 2.9.3 This time, consider a non-trivial rotational Killing field κb

=

χa ǫabcd ξ c B d on

Minkowski spacetime (with B a 6= 0). Again, let η a be a constant field on Minkowski spacetime. Show

that £κ η a = 0 iff η a is a linear combination of ξ a and B a .

The argument is very much the same as with the preceding problem. If η a is constant, £κ η a

=

− η n ∇n κa = − η n ∇n (χm ǫma cd ξ c B d )

=

− ǫma cd ξ c B d η n ∇n χm = − ǫma cd ξ c B d η n gnm = ǫa mcd η m ξ c B d .

Thus £κ η a = 0 iff η a has no component orthogonal to both ξ a and B a . Problem 2.9.4 Let κa be a Killing field; let γ : I → M be a smooth, future-directed, timelike curve,

with unit tangent field ξ a ; and let J = (P a κa ), where P a = m ξ a . Finally, let αa = ξ n ∇n ξ a and 1

α = (−αn αn ) 2 . Show that

|ξ n ∇n J| ≤ α We have seen that

p J 2 − m2 (κn κn ).

ξ n ∇n J = m κa ξ n ∇n ξ a = m κa αa . Now consider the projected spatial metric hab = gab − ξa ξb . It is negative definite. So by the Schwarz

inequality (as applied to −hab ) and the fact that ξ a αa = 0, |ξ n ∇n J| = =

1

1

|m κa αa | = |m hab αa κb | ≤ (−hab αa αb ) 2 (−m2 hab κa κb ) 2 1

α [J 2 − m2 (κn κn )] 2 .

Problem 2.11.1 Confirm that the three stated solutions do, in fact, satisfy (2.11.18). We consider just the k = −1 case. The others are handled similarly. We have to show that the solution

(in parametric form),

a(x)

=

C (cosh x − 1) 2

t(x)

=

C (sinh x − x) 2

291


does, in fact, satisfy (2.11.18) for all x ∈ (0, ∞). Note that (dt/dx) is strictly positive in this interval. So by the inverse function theorem, (dx/dt) is everywhere well defined and equal to (dt/dx)−1 . Thus, we

have

da da = a˙ = dt dx

Therefore C a˙ − −1 a 2

=

dt dx

sinh x cosh x − 1

−1 2

=

−

sinh x . cosh x − 1

2 − 1 = 0. (cosh x − 1)

Problem 2.11.2 Consider a second equation of state, namely that in which ρ = 3 p. Show that in this case there is a number C ′ such that a˙ 2 a2 + k a2 =

8π 4 ρ a = C ′. 3

If we multiply the right side of (2.11.16) by 3, and equate it with the right side of (2.11.15), we arrive at

a˙ ä k a˙ k ( )2 + 2 = −2 − ( )2 − 2 a a a a a

or, equivalently, ä a + a˙ 2 + k = 0. It follows (by integration) that a˙ 2 a2 + k a2 = C ′ , for some number C ′ . It then further follows from (2.11.15) that C ′ = (8 π/3) ρ a4 .

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